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ADVANCED  ALGEBRA 


BY 

HEEBEET  E.  HAWKES,  Ph.D. 

Assistant  Professor  of  Mathematics  in  Yale  University 


X 


GINN  &  COMPANY 

BOSTON  .  NEW  YORK  •  CHICAGO  •  LONDON 


QA\54 


e. 


0 


Copyright,  1905,  by 
H.  E.  HAWKES 


ALL  BIGHTS   KESERVED 

613.1 


GINN   &  COMPANY  •   PRO- 
PRIETORS .  BOSTON  .  U.S.A. 


PEEFACE 

This  book  is  designed  for  use  in  secondary  schools  and  in 
short  college  courses.  It  aims  to  present  in  concise  but  clear 
form  the  portions  of  algebra  that  are  required  for  entrance  to 
the  most  exacting  colleges  and  teclinical  schools. 

The  chapters  on  algebra  to  quadratics  are  intended  for  a 
review  of  the  subject,  and  contain  many  points  of  view  that 
should  be  presented  to  a  student  after  he  has  taken  a  first 
course  on  those  topics.  Throughout  the  book  the  attention 
is  concentrated  on  subjects  that  are  most  vital,  pedagogically 
and  practically,  while  topics  that  demand  a  knowledge  of  the 
calculus  for  their  complete  comprehension  (as  multiple  roots, 
and  Sturm's  theorem)  or  are  more  closely  related  to  other  por- 
tions of  mathematics  (as  theory  of  numbers,  and  series)  have 
been  omitted. 

The  chapter  on  graphical  representation,  has  been  intro- 
duced early,  in  the  belief  that  the  illumination  which  it  affords 
greatly  enlivens  the  entire  presentation  of  algebra.  The  dis- 
cussion of  the  relation  between,  pairs  of  linear  ftguntioT^^g  and 
pairs  of  straight  lines  is  particularly  suggestive.  _ 

In  each  chapter  the  discussion  is  directed  toward  a  definite 

result.     The  chapter  on  theory  of  equations  aims  to  give  a 

simple  and  clear  treatment  of  the  method  of  obtaining  the 

real  roots  of  an  equation  and  the  theorems  that  lead  to  that 

iii 


383513 


vi  CONTENTS 

CHAPTER  n 
FACTORING 

SBOnOK  ^  PAGE 

28.  Statement  of  the  Problem 16 

29.  Monomial  Factors 16 

30.  Factoring  by  grouping  Terms 17 

31.  Factors  of  a  Quadratic  Trinomial 18 

32.  Factoring  the  Difference  of  Squares 20 

33.  Reduction  to  the  Difference  of  Squares 20 

34.  Replacing  a  Parenthesis  by  a  Letter 21 

35.  Factoring  Binomials  of  the  Form  a"  ±  6« 22 

36.  Highest  Common  Factor 22 

S'7.  H.C.F.  of  Two  Polynomials 23 

i^.:)i}adid"s  Method  of  finding  the  H.C.F 23 

39.  Method  of  finding  the  H.C.F.  of  Two  Polynomials       ...  24 

40.  Least  y^o>nmon  Multiple 26 

41.  Second  Rule  for  finding  the  Least  Common  Multiple   ...  26 


CHAPTER   III 
FRACTIONS 

42.  General  Principles    . 27 

43.  Principle  I 27 

44.  Principle  II 27 

45.  Principle  III 27 

46.  Reduction.         . 27 

47.  Least  Common  Denominators  of  Several  Fractions      ...  28 

48.  Addition  of  Fractions 29 

49.  Subtraction  of  Fractions 29 

60.  Multiplication  of  Fractions 29 

51.  Division  of  Fractions 29 


CHAPTER   IV 
EQUATIONS 

52.  Introduction      . 32 

63.  Identities  and  Equations  of  Condition 32 

64.  Linear  Equations  in  One  Variable 33 

65.  Solution  of  Problems .        .  37 

66.  Linear  Equations  in  Two  Variables 40 

57.  Solution  of  a  Pair  of  Equations 40 


CONTENTS 


vu 


SECTION 

68.  Independent  Equations 

69.  Solution  of  a  Pair  of  Simultaneous  Linear  Equations 

60.  Incompatible  Equations 

61.  R^um6 

62.  Solution  of  Problems  involving  Two  Unknowns 

63.  Solution  of  Linear  Equations  in  Several  Variables 


PAGE 
.       41 

•  42 
.     42 

43 
•     45 

47 


CHAPTER   V 
RATIO  AND  PROPORTION 

64.  Ratio 49 

65.  Proportion     . 49 

66.  Theorems  concerning  Proportion 49 

67.  Theorem v  50 

68.  Mean  Proportion 60 


CHAPTER   VI 


IRRATIONAL  NUMBERS  AND  RADICALS 


69.  Existence  of  Irrational  Numbers  . 

70.  The  Practical  Necessity  for  Irrational  Numbers 

71.  Extraction  of  Square  Root  of  Polynomials   . 

72.  Extraction  of  Square  Root  of  Numbers     . 

73.  Approximation  of  Irrational  Numbers  . 

74.  Sequences 

75.  Operations  on  Irrational  Numbere 

76.  Notation 

77.  Other  Irrational  Numbers      .... 

78.  Reduction  of  a  Radical  to  its  Simplest  Form 

79.  Addition  and  Subtraction  of  Radicals  . 

80.  Multiplication  and  Division  of  Radicals   . 

81.  Rationalization 

82.  Solution  of  Equations  involving  Radicals 


62 
53 
53 
54 
55 
56 
56 
57 
57 
68 
69 
60 
61 
63 


CHAPTER   VII 
THEORY  OF  INDICES 

83.  Negative  Exponents 66 

84.  Fractional  Exponents 66 

85.  Further  Assumptions 67 

86.  Theorem 67 

87.  Operations  with  Radical  Polynomials 69 


viii  CONTENTS 


QUADRATICS   AND   BEYONI? 


i    CHAPTER   VIII 
QUADRATIC  EQUATIONS 

SECTION  PAGE 

88.  Definition 70 

89.  Solution  of  Quadratic  Equations 70 

90.  Pure  Quadratics 72 

91.  Solution  of  Quadratic  Equations  by  Factoring     ....  75 

92.  Solution  of  an  Equation  by  Factoring 75 

93.  Quadratic  Form 77 

94.  Problems  solvable  by  Quadratic  Equations 79 

95.  Theorems  regarding  Quadratic  Equations 82 

96.  Theorem 83 

97.  Theorem 84 

98.  Nature  of  the  Roots  of  a  Quadratic  Equation 84 


CHAPTER   IX 
GRAPHICAL  REPRESENTATION 

99.  Representation  of  Points  on  a  Line 87 

100.  Cartesian  Coordinates 88 

101.  The  Graph  of  an  Equation 90 

102.  Restriction  to  Coordinates 91 

103.  Plotting  Equations 91 

104.  Plotting  Equations  after  Solution 93 

105.  Graph  of  the  Linear  Equation  . "94 

106.  Method  of  plotting  a  Line  from  its  Equation        ....  96 

107.  Solution  of  Linear  Equations,  and  the  Intersection  of  their  Graphs  97 

108.  Graphs  of  Dependent  Equations 99 

109.  Incompatible  Equations 09 

110.  Graph  of  the  Quadratic  Equation 100 

111.  Form  of  the  Graph  of  a  Quadratic  Equation     ....  101 

112.  The  Special  Quadratic  ox^  +  to  =  0 103 

113.  The  Special  Quadratic  aa;^  +  c  =  0 104 

114.  Degeneration  of  the  Quadratic  Equation 104 

115.  Sum  and  Difference  of  Roots 106 

116.  Variation  in  Sign  of  a  Quadratic 107 


CONTENTS  ix 


CHAPTER  X 
SIMULTANEOUS  QUADRATIC  EQUATIONS  IN  TWO  VARIABLES 

SECTION  PAGE 

117.  Solution  of  Simultaneous  Quadratics Ill 

118.  Solution  by  Substitution Ill 

119.  Number  of  Solutions 113 

120.  Solution  when  neither  Equation  is  Linear 114 

121.  Equivalence  of  Pairs  of  Equations 120 

122.  Incompatible  Equations 121 

123.  Graphical  Representation  of  Simultaneous  Quadratic  Equations  122 

124.  Graphical  Meaning  of  Homogeneous  Equations        .         .         .  123 

CHAPTER   XI 
MATHEMATICAL  INDUCTION 

126.  General  Statement 125 

CHAPTER   XII 
BINOMIAL  THEOREM 

126.  Statement  of  the  Binomial  Theorem 128 

127.  Proof  of  the  Binomial  Theorem 129 

128.  General  Term 129 

CHAPTER   XIII 

ARITHMETICAL   PROGRESSION 

129.  Definitions :        ....  133 

130.  The  nth  Term 133 

131.  The  Sum  of  the  Series 134 

132.  Arithmetical  Means 134 

CHAPTER   XIV 

GEOMETRICAL   PROGRESSION 

133.  Definitions 137 

134.  The  nth  Term 137 

135.  The  Sum  of  the  Series 138 

136.  Geometrical  Means 138 

137.  Infinite  Series 140 


CONTENTS 


ADVANCED   ALGEBEA 


CHAPTER  XV 

PERMUTATIONS  AND   COMBINATIONS 
SECTION  PAGE 

138.  Introduction 143 

139.  Permutations 144 

140.  Combinations 146 

141.  Circular  Permutations 149 

142.  Theorem 160 

CHAPTER  XVI 

COMPLEX  NUMBERS 

143.  The  Imaginary  Unit 152 

144.  Addition  and  Subtraction  of  Imaginary  Numbers     .         .         .  163" 

145.  Multiplication  and  Division  of  Imaginaries 154 

146.  Complex  Numbers 156 

147.  Graphical  Representation  of  Complex  Numbers    ....  155 

148.  Equality  of  Complex  Numbers  . 155 

149.  Addition  and  Subtraction 156 

150.  Graphical  Representation  of  Addition 156 

151.  Multiplication  of  Complex  Numbers 157 

152.  Conjugate  Complex  Numbers    .         .         .         .         .         .         .  158 

153.  Division  of  Complex  Numbers 158 

154.  Polar  Representation 160 

155.  Multiplication  in  Polar  Form 160 

156.  Powers  of  Numbers  in  Polar  Form 161 

157.  Division  in  Polar  Form 162 

158.  Roots  of  Complex  Numbers 162 

CHAPTER   XVII 


THEORY  OF  EQUATIONS 

159.  Equation  of  the  nth  Degree 166 

160.  Remainder  Theorem 166 

161.  Synthetic  Division .        .  167 

162.  Proof  of  the  Rule  for  Synthetic  Division 169 

163.  Plotting  of  Equations    .         . 170 

164.  Extent  of  the  Table  of  Values 171 


^ 


CONTENTS  xi 

SIECTION  PAGE 

fl65.  Roots  of  an  Equation 172 

.  Number  of  Roots 172 

167.  Graphical  Interpretation 174 

^  168.  Imaginary  Roots .         .         .  174 

.  Graphical  Interpretation  of  Imaginary  Roots       ....  175 

170.  Relation  between  Roots  and  Coefficients 177 

171.  The  General  Term  in  the  Binomial  Expansion      ....  178 

172.  Solution  by  Trial .  178 

173.  Properties  of  Binomial  Surds 179 

174.  Formation  of  Equations .         .  180 

176.  To  multiply  the  Roots  by  a  Constant 183 

176.  Descartes'  Rule  of  Signs 186 

177.  Negative  Roots 189 

178.  Integral  Roots 190 

179.  Rational  Roots 190 

180.  Diminishing  the  Roots  of  an  Equation       .         .         ...         .  191 

181.  Graphical  Interpretation  of  Decreasing  Roots       ....  193 

182.  Location  Principle 194 

183.  Approximate  Calculation  of  Roots  by  Horner's  Method       .        .  195 

184.  Roots  nearly  Equal 200 


CHAPTER  XVIII 

DETERMINANTS 

185.  Solution  of  Two  Linear  Equations 203 

186.  Solution  of  Three  Linear  Equations 204 

187.  Inversion 208 

188.  Development  of  the  Determinant 208 

189.  Number  of  Terms 210 

190.  Development  by  Minors 210 

191.  Multiplication  by  a  Constant 213 

192.  Interchange  of  Rows  and  Columns 213 

193.  Interchange  of  Rows  or  Columns 214 

194.  Identical  Rows  or  Columns        .        .        .        .        .        .        .  215 

195.  Proof  for  Development  by  Minors 215 

196.  Sum  of  Determinants 216 

197.  Vanishing  of  a  Determinant 217 

198.  Evaluation  by  Factoring 218 

199.  Practical  Directions  for  evaluating  Determinants  .         ...  219 

200.  Solution  of  Linear  Equations 221    ^ 

201.  Solution  of  Homogeneous  Linear  Equations  ....  223  4^^ — -• 


xii  CONTENTS 

CHAPTER  XIX 

PARTIAL   FRACTIONS 
SECTION  .^  PAGE 

202.  Introduction 226 

203.  Development  when  (f){x)=  0  has  no  Multiple  Roots      .         .         .  225 

204.  Development  when  (p{x)=0  has  Imaginary  Roots    .         .         .  229 

205.  Development  when  <p  (x)  =  {x  —  ar)» 232 

206.  General  Case 233 

CHAPTER   XX 
LOGARITHMS 

207.  Generalized  Powers 235 

208.  Logarithms         .  ' 236 

209.  Operations  on  Logarithms 237 

210.  Common  System  of  Logarithms 239 

211.  Use  of  Tables 241 

212.  Interpolation 242 

213.  Antilogarithms 243 

214.  Cologarithms 246 

215.  Change  of  Base 247 

216.  Exponential  Equations 261 

217.  Compound  Interest 263 

CHAPTER   XXI 
CONTINUED  FRACTIONS 

218.  Definitions          .       ' 256 

219.  Terminating  Continued  Fractions 256 

220.  Convergents 258 

221.  Recurring  Continued  Fractions 260 

222.  Expression  of  a  Surd  as  a  Recurring  Continued  Fraction          .  263 

223.  Properties  of  Convergents 266 

224.  Limit  of  Error 267 

CHAPTER   XXII 
INEQUALITIES 

225.  General  Theorems .        .269 

226.  Conditional  Linear  Inequalities 271 

227.  Conditional  Quadratic  Inequalities 271 


CONTENTS  xiii 


CHAPTER   XXm 
VARIATION 

SECTION  ~  PAOB 

228.  General  Principles .  .273 


CHAPTER  XXIV 
PROBABILITY 

229.  Illustration 276 

230.  General  Statement 276 

CHAPTER  XXV 
SCALES   OF  NOTATION 

231.  General  Statement 279 

232.  Fundamental  Operations  ...  ....  280 

233.  Change  of  Scale     .....  ....  281 

234.  Fractions 282 

236.  Duodecimals  .        ,  284 


ik 


ADVANCED  ALGEBRA 


ALGEBRA  TO   QUADRATICS 

CHAPTER  I 
FUNDAMENTAL  OPERATIONS 

1.  It  is  assumed  that  the  elementary  operations  and  the  mean- 
ing of  the  -usual  symbols  of  algebra  are  familiar  and  do  not 
demand  detailed  treatment.  In  the  following  brief  exposition 
of  the  formal  laws  of  algebra  most  of  the  proofs  are  omitted. 

2.  Addition.  The  process  of  adding  two  positive  integers  a  and 
h  consists  in  finding  a  number  x  such  that 

a  -\-h  =  x. 

For  any  two  given  positive  integers  a  single  sum  x  exists 
which  is  itself  a  positive  integer. 

3.  Subtraction.  The  process  of  subtracting  the  positive  num- 
ber h  from  the  positive  number  a  consists  in  finding  a  number  x 

such  that 

h  -\-  x  =  a.  (1) 

This  number  x  is  called  the  difference  between  a  and  h  and  is 

denoted  as  follows : 

a  —  6  =  a;, 

a  being  called  the  minuend  and  h  the  subtrahend. 

If  a  >  ft  and  both  are  positive  integers,  then  a  single  posi- 
tive integer  x  exists  which  satisfies  the  condition  expressed  by 
equation  (1) 

1 


. AL'G^BE a:  to  *  ^JJ ADR ATI€S  ^ 


li  a  <h,  then  x  is  not  a  positive  integer.  In  order  that  the  pro- 
cess of  subtraction  may  be  possible  in  this  case  also,  we  introduce 
negative  numbers  which  we  symbolize  by  (—a  ),  (—  b),  etc.  When 
in  the  difference  a~b,ais  less  than  b,  we  define  a  —  b  =  (—  (b  —  a)). 
The  processes  of  addition  and  subtraction  for  the  negative  numbers 
are  defined  as  follows : 

a  -^  (—  b)  =  a  —  b. 

a  —  (—  b)  =  a  -^  b. 

4.  Zero.  If  in  equation  (1),  a  =  b,  there  is  no  positive  or  nega- 
tive number  which  satisfies  the  equation.  In  order  that  in  this 
case  also  the  equation  may  have  a  number  satisfying  it,  we  intro- 
duce the  number  zero  which  is  symbolized  by  0  and  defined  by 

the  equation 

a  -\-  0  =  a, 

or  a  —  a  =  0. 

The  processes  of  addition  and  subtraction  for  this  new  number 
zero  are  defined  as  follows,  where  a  stands  for  either  a  positive  or 
a  negative  number 

0-a  =  -a. 
0  ±  0  =  0. 

5.  Multiplication.  The  process  of  multiplying  a  by  i  consists 
in  finding  a  number  x  which  satisfies  the  equation 

a-b  =  X. 

*  The  symbol  for  a  positive  integer  might  be  written  (+  a),  (+  6),  etc.,  consistently 
with  the  notation  for  negative  numbers.  Since,  however,  no  ambiguity  results,  we  omit 
the  +  sign.  Since  the  laws  of  combining  the  +  and  —  signs  given  in  this  and  the  folloM'ing 
paragraphs  remove  the  necessity  for  the  parentheses  in  the  notation  for  the  negative 
number,  we  shall  omit  them  where  no  ambiguity  results. 


FUNDAMENTAL  OPERATIONS  3 

When  a  and  h  are  positive  integers  cc  is  a  positive  integer  which, 
may  be  found  by  adding  a  to  itseK  h  times.  When  the  numbers 
to  be  multiplied  are  negative  we  have  the  following  laws, 

(—  a)  •  (—  5)  =  a  •  5, 

Oa  =  «;.0  =  0,  (1) 

where  «;  is  a  positive  or  negative  number  or  zero. 

These  symbolical  statements  include  the  statement  of  the 
following 

Principle.  A  'product  of  numbers  is  zero  when  and  only  when 
one  or  more  of  the  factors  are  zero. 

This  most  important  fact,  which  we  shall  use  continually,  assures 
us  that  when  we  have  a  product  of  several  numbers  as 

a-b-G-  d  =^  Bj 

first,  if  e  equals  zero,  it  is  certain  that  one  or  more  of  the  num- 
bers a,  bj  c,  or  d  are  zero ;  second,  if  one  or  more  of  the  numbers 
a,  b,  c,  or  d  are  zero,  then  e  is  also  zero. 

6.  Division.  The  process  of  dividing  ahj  p  consists  in  finding 
a  number  x  which  satisfies  the  equation 

X-p  =  a,       .  (1) 

where  a  and  jS  are  positive  or  negative  integers,  or  a  is  0. 
When  a  occurs  in  the  sequence  of  numbers 

•  ••-.?A  -2A  -Ap,  ft-2A  3A  •••, 

a;  is  a  definite  integer  or  0,  that  is,  it  is  a  number  such  as  we 
have  previously  considered.  If  a  is  not  found  in  this  series,  but 
is  between  two  numbers  of  the  series,  then  in  order  that  in  this 
case  the  process  may  also  be  possible  we  introduce  the  fraction 

which  we  symbolize  by  a  -*-  /8  or  -  and  which  is  defined  by  the 
equation 


4  ALGEBRA  TO  QUADRATICS 

The  operations  for  addition,  subtraction,  multiplication,  and  divi- 
sion of  fractions  are  defined  as  follows : 

fi       8       Py  W 

Further  properties  of  fractions  are  the  following : 

a      1 


—  =  —  >    where  S  is  any  number,  (5) 


j8     SP 

The  last  two  equations  are  expressed  verbally  as  follows : 

Both  numerator  and  denominator  of  a  fraction  may  be  multi- 
plied by  any  number  without  changing  the  value  of  the  fraction. 

Changing  the  sign  of  either,  numerator  or  denominator  of  a 
fraction  is  equivalent  to  changing  the  sign  of  the  fraction. 

The  laws  of  signs  in  multiplication  given  on  p.  3  may  now  be 
assumed  to  hold  when  the  letters  represent  fractions  as  well  as 
integers.  j-      ,.-,     . 

Thus  for  example     ~  I  t  )     '     "~ 


ac 
M 


The  positive  or  negative  number  a  may  be  written  in  the 

fractional  form 

a 

1* 

7.  Division  by  zero.  If  in  equation  (1),  ^Q>^  p=i  0,  there  is  no 
single  number  x  which  satisfies  the  equation,  since  by  (1),  §  5, 
whatever  value  x  might  have,  its  product  with  zero  would 
be  zero. 


FUNDAMENTAL  OPERATIONS  5 

Thus  division  by  zero  is  entirely  excluded  from  algebraic  pro- 
cesses. Before  a  division  can  safely  be  performed  one  must  be 
assured  that  the  divisor  cannot  vanish.     In  the  equation 

4  0  =  2.0, 

if  we  should  allow  division  of  both  sides  of  our  equation  by  zero, 
we  should  be  led  to  the  absurd  result  4  =  2. 

8.  Fundamental  operations.  The  operations  of  addition,  sub- 
traction, multiplication,  and  division  we  call  the  four  fundamental 
operations.  Any  numbers  that  can  be  derived  from  the  number  1 
by  means  of  the  four  fundamental  operations  we  call  rational  num- 
bers. Such  numbers  comprise  all  positive  and  negative  integers 
and  such  fractions  as  have  integers  for  numerator  and  denominator. 
Positive  or  negative  integers  are  called  integral  numbers. 

9.  Practical  demaild  for  negative  and  fractional  numbers. 
In  the  preceding  discussion  negative  numbers  and  fractions  have 
been  introduced  on  account  of  the  mathematical  necessity  for 
them.  They  were  needed  to  make  the  four  fundamental  opera- 
tions always  possible.  That  this  mathematical  necessity  corre- 
sponds to  a  practiqal  necessity  appears  as  soon  as  we  attempt 
to  apply  our  four  operations  to  practical  affairs.  Thus  if  on  a 
certain  day  the  temperature  is  +  20°  and  the  next  day  the  mer- 
cury falls  25°,  in  order  to  express  the  second  temperature  we 
must  subtract  2^  from  20.  If  we  had  not  introduced  negative 
numbers,  this  would  be  impossible  and  our  mathematics  would 
be  inapplicable  to  this  and  countless  other  everyday  occurrences. 

10.  Laws  of  operation.  All  the  numbers  which  we  use  in 
algebra  are  subject  to  the  following  laws. 

' .Commutative  law  of  addition.  This  law  asserts  that  the  value 
of  the  sum  of  two  numbers  does  not  depend  on  the  order  of 
summation. 

Symbolically  expressed, 

a  -\-  h  =  h  -\-  aj 

where  a  and  h  represent  any  numbers  such  as  we  have  presented 
or  shall  hereafter  introduce. 


6  ALGEBRA  TO  QUADRATICS 

Associative  law  of  addition.  This  law  asserts  that  the  sum  of 
three  numbers  does  not  depend  on  the  way  in  which  the  numbers 
are  grouped  in  performing  the  process  of  addition. 

Symbolically  expressed, 

a  4-  (^  +  c)  =  (a  +  &)  +  c  =  a  4-  *  +  c. 

Commutative  law  of  multiplication.  This  law  asserts  that  the 
value  of  the  product  of  two  numbers  does  not  depend  on  the  order 
of  multiplication. 

Symbolically  expressed, 

a-b  =  b-a. 

Associative  law  of  multiplication.  This  law  asserts  that  the 
value  of  the  product  of  three  numbers  does  not  depend  on  the  way 
in  which  the  numbers  are  grouped  in  the  process  of  multiplication. 

Symbolically  expressed, 

(^a-b)  -c  =  a-(b'C)=  a-b-c. 

Distributive  law.  This  law  asserts  that  the  product  of  a  single 
number  and  the  sum  of  two  numbers  is  identical  with  the  sum 
of  the  products  of  the  first  number  and  the  other  two  numbers 
taken  singly. 

Symbolically  expressed, 

a-  (b  -\-  c)=  a-b  -{-  a-c. 

All  the  above  laws  are  readily  seen  to  hold  when  more  than  three  numbers  are 
involved. 

11.  Integral  and  rational  expressions.  A  polynomial  is  inte- 
gral when  it  may  be  expressed  by  a  succession  of  literal  terms,  no 
one  of  which  contains  any  letter  in  the  denominator. 

Thus  4  a;5  _  a;3  _  2  x2  _  J  x  +  1  is  integral. 

The  quotient  of  two  integral  expressions  is  called  rational. 

^^      a;2_2a;  +  3.       ^.       , 
Thus z —  18  rational. 

a  — 7 

12.  Operations  on  polynomials.  We  assume  that  the  same 
formal  laws  for  the  four  fundamental  operations  enunciated  in 
§  §  2-6  and  the  laws  given  in  §  10  hold  whether  the  letters  in  the 
symbolic  statements  represent  numbers  or  polynomials. 


FUNDAMENTAL  OPERATIONS  7 

In  fact  the  literal  expressions  which  we  use  are  in  essence 
nothing  else  than  numerical  expressions,  since  the  letters  are  merely 
symbols  for  numbers.  When  the  letters  are  replaced  by  numbers, 
the  literal  expressions  reduce  to  numerical  expressions  for  which 
the  previous  laws  have  been  explicitly  given. 

13.  Addition  of  polynomials.  For  performing  this  operation 
we  have  the  following 

EuLE.    Write  the  terms  with  the  same  literal  part  in  a  column. 
Find  the  algebraic  sum  of  the  terms  in  each  column,  and  write 
the  results  in  succession  with  their  proper  signs. 

When  the  polynomials  reduce  to  monomials  the  same  rule  is  to  be  observed. 

EXERCISES 

Add  the  following : 

1.  3a262_2a6  +  6a26-a;  4.  ab  -  2  a^h'^  -  11  a'^b  +  9  a  ; 

2a^b -ab -2a;  Sam- 4:db -6a. 

Solution:  ZaW-2ab+   6a^b-    a 

-2a262  +  4a6-  lla26  +  9a 

-    a6  +   2  a26  -  2  a 

Sa^-iab  -6a 

9am -Sab-   Sa^b 

2.  21a-246-8c2;  16c  +  17  6  +  6c2  -  20a  ;  186 -18c. 

3.  x*  -6a;2-  8x-l;  2x3  +  1;  6x'^  +  1  x  +  2  ;  x^  -  x^  +  x  -  1. 

4.  9(a  +  6)-6(6  +  c)  +  7(a  +  c);  4(6  + c)  -  7(a  +  6)  -  8(a  +  c)  ,• 

(a  +  c)  -  (a  +  6)  +  (6  +  c). 

5.  a2-4a6  +  62  +  a  +  6-2;  2a2  +  4a6-  362-  2a -26 +  4; 

3a2-  5a6-462  +  3a  +  36  -  2;  6a2  +  10a6  +  562  +  a  +  6. 

14.  Subtraction  of  polynomials.  For  performing  this  opera- 
tion we  have  the  following 

EuLE.  Write  the  subtrahend  under  the  minuend  so  that  terms 
with  the  same  literal  part  are  in  the  same  column. 

To  each  term  of  the  minuend  add  the  corresponding  term  of 
the  subtrahend,  the  sign  of  the  latter  having  been  changed. 

It  is  generally  preferable  to  imagine  the  signs  of  the  subtrahend  changed  rather 
than  actually  to  write  it  with  the  changed  signs.  


8  ALGEBRA  TO  QUADRATICS 

EXERCISES 

1.  From  a262  -  3  a^ft  +  8  a6  +  6  6  subtract  9  a'^h'^  -  6  a6  +  4  a^ft  +  a. 

Solution :  a%'^  -  3  a26  +    8  a&  +  6  6 

-9a262:f  4a26-    6a6  +a 

-  8a262  ^  7 a26  +  14a6  +  66  -  a 

2.  From  6  ahx  —  4  win  +  5  x  subtract  3  mn  +  6  ax  —  4  a6x. 

3.  From  m  -{■  an  +  bq  subtract  the  sum  of 

cm  +  dn  +  {b  —  a)q  and  {a  —  b)q  —  {a  -\-  d)n  —  cm. 

4.  From  the  sum  of  |  a  +  y^  6  +  |  c  and  —  6  —  c  —  a  subtract  ^6  —  |c  +  ^a. 

5.  From  the  sum  of  2  x^  —  3  x  +  4  and  x*  —  f  x  —  i  subtract  x^  —  |  x^ 

-3|x  +  3i. 

15.  Parentheses.  When  it  is  desirable  to  consider  as  a  single 
symbol  an  expression  involving  several  numbers  or  symbols  for 
numbers,  the  expression  is  inclosed  in  a  parenthesis.  This  paren- 
thesis may  then  be  used  in  operations  as  if  it  were  a  single  number 
or  symbol,  as  in  fact  it  is,  excepting  that  the  operations  inside 
the  parenthesis  may  not  yet  have  been  carried  out. 

EuLE.  When  a  single  parenthesis  is  preceded  hy  a  -\-  sign 
the  parenthesis  may  he  removed,  the  various  terms  retaining 
the  same  sign. 

When  a  single  parenthesis  is  preceded  hy  a  —  sign  the  parent- 
thesis  may  be  removed,  providing  we  change  the  signs  of  all  the 
terms  inside  the  parenthesis. 

When  several  parentheses  occur  in  an  expression  we  have  the 
following 

EuLE.  Remove  the  innermost  parenthesis,  changing  the  signs 
of  the  terms  inside  if  the  sign  preceding  it  is  minus. 

Simplify,  if  possible,  the  expression  inside  the  new  inn^rwjost 
parenthesis. 

Repeat  the  process  until  all  the  parentheses  are  removed. 

It  is  in  general  unwise  to  shorten  the  process  by  carrying  out  some  of  the  steps 
in  one's  head.  The  liability  to  error  in  such  attempts  more  than  offsets  the  gain 
in  time. 


FUNDAMENTAL  OPERATIONS 


EXERCISES 

Remove  parentheses  from  the  following : 


1.  6  _  pa -[26  + (4a -2a -6) -66]}. 


Solution:  6  -  {9a  -  [26  +  (4a  -  2a  -  6)  -  66]} 

=  6- {9a -[26  + (4a -2a +  6) -66]} 
=  6-{9a-[2  6+  (2a  +  6)-6  6]} 
=  6  _  [9  a  -  (2  6  +  2  a  +  6  -  6  6)] 
=  6- [9a -(2a -36)] 
=  6 -(9a -2a +  36) 
=  6  -  (7  a  +  3  6) 
=  6-7a-36 
=  -  7  a  -  2  6. 

2-  -{-[-[-(-(-I))]]}- 

3.  a2  +  4  -  {6  -  [-  (a2  -  6)  +  1]}. 

4.  i{i-|[f-ia-i-f-T\)  +  |]-^}- 

5.  x2  -  {2/2  _  [4x  +  3(y  -  9x(y  -  x))  +  9y  (x  -  y)]}. 

6.  Find  the  value  of  a  -  {5 6  -  [a  -  (3 c  -  3 6)  +  2 c  -  3 (a  -  26 -c)]}     i^ 
when  a  =  —  3,  6  =  4,  c  =  —  5. 

16.  Multiplication.    It    is    customary    to    write    a  -  a  =  a^ ; 
aaa  =  a^]  a  -  a  •   •  a  =  a"*.    We  have  tlien  by  tlie  associative 

n  terms 

law  of  multiplication,  §  10, 

a^  -  a^  =  (a  •  a)  (a  -  a  '  a)  =  a   a  •  a  '  a  •  a  =  a^, 
or,  in  general. 


where  m  and  n  are  positive  integers. 
Furthermore, 


(I) 

(a')"  =  a"-a'---a'  =  a'-"'.  (II) 


m  terms 


Finally,  a"  •  J"  =  (a  •  bf.  (Ill) 

The  distinction  between  (a**)"*  and  a""*  should  be  noted  carefully.  Thus  (28)* 
=  82  =  64,  while  28*  =2^  =  512. 

Equation  (I)  asserts  that  the  exponent  in  the  product  of  two 
powers  of  any  expression  is  the  sum  of  the  exponents  of  the 
factors.    Hence  we  may  multiply  monomials  as  follows : 


10  ALGEBRA  TO  QUADRATICS 


I 


Rule.    Write  the  product  of  the  numerical  coefficients,  followed  , 
by  all  the  letters  that  occur  in  the  multiplier  and  multiplicand, 
each  having  as  its  exponent  the  sum  of  the  exponents  of  that 
letter  in  the  multiplier  and  multiplicand. 

Example.    4  a2^i0c#  •  (-  16  a'^bd'^)  =  -  64  a^b^^cd^\ 

17.  Multiplication   of   monomials   by   polynomials.   By  the 

distributive  law,  §  10,  we  can  immediately  formulate  the 

EuLE.  Multiply  each  term  of  the  polynomial  by  the  monomial 
and  write  the  resulting  terms  in  succession. 

Example.  9  a'^h'^  -  2  a&  +  4  a62  -  a  +  &* 

3a26 

27  a463  _  6  aW  +  12  a^h^  -  3  a^fe  +  3  aW 
* 

18.  Multiplication  of  polynomials.    If  in  the  expression  for  the 

distributive  law,  a(c  +  d)=  ac  •{- ad, 

we  replace  a  by  a  +  5,  we  have 

(a  -\-  h)  (c  -\-  d)  =  ac  -\- be  -\-  ad  -{-  bd, 
which  affords  the 

EuLE.  Multiply  the  mult^icand  by  each  term  of  the  multi- 
plier in  turn,  and  write  the  partial  products  in  succession. 

To  test  the  accuracy  of  the  result  assume  some  convenient  numer- 
ical value  for  each  letter,  and  find  the  corresponding  numerical 
value  of  multiplier,  multiplicand,  and  product.  The  latter  should 
be  the  product  of  the  two  former. 

EXERCISES 


1.  Multiply  and  check  the  following: 

(a)  2  a2  +  a6  +  4  62  +  5  and  a-h-\-  ah. 

Check : 

Solution : 

Let  a  =  6  =  1 

2a2+     a6+462    +    6 

=  8 

a    -    h     +  a6 

=  1 

2a8+     a26  +  4a&2  +  a6 

-2a26-     a62           -463_ 

-62 

+    a62 

+  2a86  +  a262  +  4a68 

2a8  -     a26  +  4a62  +  a6  -  46^  -  62  -f-  2a86  +  a262  +  4a68  =  8 


FUNDAMENTAL  OPERATIONS  H 

(b)  6  abx^  and  4  a'^b^x. 

(c)  ^^and  -OxV^. 

o 

(d)  3  a62x  -  ^  6x4  and  6  a^cx. 

(e)  x2«  +  ?/26  ^  x^  and  x«  —  y^ 

(f )  a^  +  ab  +  l^  and  a^  +  ac  +  c2. 

(g)  x^y^'^j  a;»-3yin4-4^  and  ic^y2m-2^ 
(h)  xP-3  4- xP-2  +  1  and  a;3  _  a;2  _  1, 

(i)  8  a26c,  -  a62,  _  7  62,  -  —  a^c*,  and  -. 
w  '  4      '  '       14        '         6 

(j)  ax*  —  2  a2x3  —  X  +  4  a  and  —  x  +  2  a. 
(k)  x«  +  *  +  x2«  +  x2&  4-  a;3o-&  and  x«-^  —  1. 

(1)  15x*  -  11x3  +  6x2  +  2x  -  1  and  -  3x2  -  1. 
(m)  4  x*  —  8  xy^  +  |  x2y2  _  3  x^y  —  x  —  y  and  —  42  xy. 

2.  Expand  (x  +  y)*. 

3.  Expand  and  simplify 

(a;2  +  2/2  +  22)2  _  (X  +  y  +  2r)  (X  +  y  -  z)  (x  +  2  -  y)  (y  +  2  -  X). 

19.  Types  of  multiplication.  The  following  types  of  multi- 
plication should  be  so  familiar  as  merely  to  require  inspection  of 
the  factors  in  order  to  write  the  product. 

EuLE.  The  product  of  the  sum  and  difference  of  two  terms  is 
equal  to  the  square  of  the  terms  with  like  signs  minus  the  square 
of  the  terms  which  have  unlike  signs. 

Examples.  (a  -  6)  (a  +  6)  =  a?^  —  62. 

(4x2  -  3y2)(4x2  +  3y2)  =  16x4  -  9y4. 

20.  The  square  of  a  binomial.  This  process  is  performed  as 
follows : 

KuLE.  The  square  of  a  binomial,  or  expression  in  two  terms, 
is  equal  to  the  sum  of  the  squares  of  the  two  terms  plus  twice 
their  product. 

Examples.  (x  +  y)2  =  x2  +  y2  +  2  xy. 

(2a-36)2=:4a2  +  962-  12a6. 


12  ALGEBRA  TO  QUADRATICS 

21.  The  square  of  a  polynomial.   This  process  is  performe! 
as  follows : 

Rule.  The  square  of  any  polynomial  is  equal  to  the  sum 
of  the  squares  of  the  terms  plus  twice  the  product  of  each  term 
hy  each  term  that  follows  it  in  the  polynomial. 

Example,    (a  +  6  +  c)2  =  a^  +  62  +  c^  +  2  a6  +  2  ac  +  2  6c. 

22.  The  cube  of  a  binomial.  This  process  is  performed  as 
follows : 

Rule.    The  cube  of  any  binomial  is  given  hy  the  folloimng 

expression :        ^^  _^  ^y  ^  ^s  _^  ^  ^2j  -\.3ab^  +  h\ 

EXERCISES 

Perform  the  following  processes  by  inspection. 
1.  (a-6  +  c)2.  2.  (a4-66)2. 

3.  (2x'-i-l)2  4.  (a2-62)3. 

5.  (2x'-i-l)3.  6.  (l-8a;22/)2. 

7.    (X2-2X  +  1)2.  8.    (a;2  -  y2  +  ^2)2. 

9.  (2a-26-c)2.  10.  (x8-2x-l)2. 

11.  (ai'-3-6p  +  3)2.  12.  (-6x2y +  4xy2)8. 

13.  {xp  -  2/9)  {xp  +  y^).  14.  (-  6x22/  +  4  xy^)^. 

15.  (3x+ 2  2/)(3x-2  2/).  16.  (- 3ax2  +  2ax -6)2. 

17.  {-3x^y-^lz^){Sx'^  +  ^z^).  18.  (_  4  -  6a26)  (- 4  +  6a26). 

19.  (2 a -2^.  20.  (2 a-?)'. 

23.  Division.    By  the  definition  of  division  in  §  6,  we  have 

a  =  —}  a'^  =  — )   a^  =  —z'y 
.  a  a  a'' 

or,  in  general,  ^ 

a"-™  =  — -> 
a"" 

where  n  and  m  are  positive  integers  and  n>  m. 

If  n  =  m,  we  preserve  the  same  principle  and  write 

a«-»  =  a«  =  —  =  1, 


(1) 


FUNDAMENTAL  OPERATIONS  13 

24.  Division  of  monomials.  Keeping  in  mind  the  rule  of  signs 
for  division  given  in  §  6,  we  have  the  following 

EuLE.  Divide  the  numerical  coefficient  of  the  dividend  hy 
that  of  the  divisor  for  the  numerical  coefficient  of  the  quotient^ 
keeping  in  mind  the  rule  of  signs  for  division. 

Write  the  literal  part  of  the  dividend  over  that  of  the 
divisor  in  the  form  of  a  fraction^  and  perform  on  each  pair 
of  letters  occurring  in  both  numerator  and  denominator  the 
process  of  division  as  defined  hy  equation  (I)  in  the  preceding 
paragraph. 

Example.   Divide  12  a'^lA^cH  by  -  6  a^hc^d^. 
12  a^6"c2d  _      2  bio 

25.  Division  of  a  polynomial  by  a  monomial.  This  process  is 
performed  as  follows : 

Rule.    Divide  each  term  of  the  polynomial  hy  the  monomial 
and  write  the  partial  quotients  in  succession. 
Example.    Divide  8  a^lP  -  12  a^h^  by  2  aW. 

8a266  _  12 gcftg  _  463  _  6a» 

2a363       2a363  ~    a         &   * 

26.  Division  of  a  polynomial  by  a  polynomial.  This  process 
is  performed  as  follows  : 

Rule.  Arrange  hoth  dividend  and  divisor  in  descending 
powers  of  some  common  letter  {called  the  letter  of  arrangement)- 

Divide  the  first  term  of  the  dividend  hy  the  first  term  of  the 
divisor  for  the  first  term  of  the  quotient. 

Multiply  the  divisor  hy  this  first  term  of  the  quotient  and 
subtract  the  product  from  the  dividend. 

Divide  the  first  term  of  this  remainder  hy  the  first  term  of  the 
divisor  for  the  second  term  of  the  quotient,  and  proceed  as 
before  until  the  remainder  vanishes  or  is  of  lower  degree  in  the 
letter  of  arrangement  than  the  divisor. 


14  ALGEBRA  TO  QUADRATICS 

When  the  last  remainder  vanishes  the  dividend  is  exactly  divisible  by 
divisor.    This  fact  may  be  expressed  as  follows: 

dividend  _        . .     . 

,.   . =  quotient. 

divisor 

When  the  last  remainder  does  not  vanish  we  may  express  the  result  of  division 

dividend  ..     ,    ,  remainder 

-^rr-. =  quotient  -\ -— : 

divisor  divisor 

The  coefficients  in  the  quotient  will  be  rational  numbers  if  those  in  both  divi- 
dend and  divisor  are  rational. 

EXERCISES 

Divide  and  check  the  following : 

1.  8a8  +  6a26  +  9a62  +  963by4a  +  &. 

Solution :  4a4-&|  8  a^  +  6  a26  +  9  a62  +  9  h^\2  a^  +  ab +  2lfl 

8  a3  +  2  a'^b 

4  a26  +  9  a62 
4a26+    am 

8a&2  +  9&3 
8a&2  +  268 
76» 
7  63  .  ■ 

Result :   2a^-\-ah-{-2,b^  + 

1  4a  +  6    • 

Check :   Let  a  =  h  =1.    Dividend  =  32,  divisor  =  5,  quotient  =  6§. 
32  -  5  =  6f . 

2.  xi2  -  yi2  by  x3  -  2/3.  3.  2 x2  -  6x  +  2  by  X  -  2. 
4.  xi2  —  2/12  i3y  aj4  _  2/*.  5.  x^  —  2/6  by  x^  +  xy  -\-  y^. 
6.  a8  -  a2  +  2  bya  +  1.  7.   -  63x*y32;2  ^y  _  gx^y^z. 

8.  .x2  -  X  -  30  by  X  +  5.  9.  4  a26  -  6  a62  _  2  a  by  -  2a.     • 

10.  16  a264cii  by  -  2  a^¥c\  11.  i  x2  -  3^  x  -  |  by  1^  x  +  T»ff. 

12.  ax2  +  (a2  -  6)  X  -  a6  by  X  +  a. 

13.  ax«  +  6x»-i  +  cx«-2  -  dx«-3  by  x'. 

14.  16  a2x22/2  _  8  ax32/2  -  4  x*?/  by  -  f  xy. 

15.  ISai'b^  -\-  6aP  +  2?^+3_  9 ap  +  ^ft^  by  3 apft*. 

16.  a2  -  2  a6  -  4  c2  +  8  6c  -  3  62  by  a  -  2  c  +  6. 

17.  x*  —  (d  +  6  +  c)  x2  -f  {ab  +  ac  +  6c)  x  -  a6c  by  x  -  a. 

18.  2j/2  _  6x2  +  i/xy  +  V-x  -  -^y  +  1  by  2x  +  |y  -  f. 

19.  x8  -  2  x22/  -  x2  +  2/2x  f  2  xy  -  y  -  2/2  +  2  by  X  -  y  +  1. 

20.  3x8 +  6x22/ +  9x2 +  2x2/2  +  5^3  +  22/ +  6y2  +  3  by  x  +  22/ +  3. 


FUNDAMENTAL  OPERATIONS  15 

27.  Types  of  division.  The  following  types  of  division,  which 
may  be  verified  by  the  rule  just  given  for  any  particular  integral 
value  of  rij  should  be  so  familiar  that  they  may  be  performed  by 
inspection. 

(a^n  _  ^,2n)  ^  (^^»  -t  ^n^  =  «»»  ^  ^«.  (I) 

(a"  +  b^)  -^  (a  +  5)  =  a"-i  -  a^-%  +  a^-%'' +  b^-\     (II) 

where  n  is  odd. 

(a"  -.  h^)  ^(a-b)^  a"-i  4-  a"-'^  +  a«-3^>2  +  . . .  +  j«-i^  (III) 
where  n  is  odd  or  even. 

EXERCISES 

Give  by  inspection  the  results  of  the  following  divisions. 
1.  a«  -  1  by  a  -r  l:  2.  a^  +  1  by  a  +  1. 

3.  x'  +  128  by  ic  +  2.  4.  x^  +  y^hy  x-\-  y. 

6.  x^  —  y^  by  x^  -{■  y^.  6.  x^  —  y*  hj  x  —  y. 

7.  x8  -  ^8  by  X*  -  y*.  8.  a2m  _  1  by  a  - 1. 

9.  a2«+^  -  1  by  a  -  1.  10.  27a9  +  868  by  3^8  +  26. 

11.  8x8-27  by  2a:- 3.  12.  4a2  -  25668  by  2a  +  1664. 

13.  16  a*  -  256  by  4  a^  +  16.  14.  27  ai^  -  64  612  by  3  a*  -  4  6*. 


CHAPTER  II 
FACTORING 

28.  Statement  of  the  problem.  The  operation  of  division  con- 
sists in  finding  the  quotient  when  the  dividend  and  divisor  are 
given.  The  product  of  the  quotient  and  the  divisor  is  the  divi- 
dend, and  the  quotient  and  the  divisor  are  the  factors  of  the 
dividend.  Thus  the  process  of  division  consists  in  finding  a 
second  factor  of  a  given  expression  when  one  factor  is  given. 

The  process  of  factoring  consists  in  finding  all  the  factors  of 
a  polynomial  when  no  one  of  them  is  given.  This  operation  is  in 
essence  the  reverse  of  the  operation  of  multiplication.  We  shall 
be  concerned  only  with  those  factors  that  have  rational  coefficients. 

29.  Monomial  factors.    By  the  distributive  law,  §  10, 

ah  -{•  ac  =  a(})  -\-  c). 
This  affords  immediately  the 

Rule.  Write  the  largest  monomial  factor  which  occurs  in  every 
term  outside  a  jpareiithesis  which  includes  the  algebraic  sum  of 
the  remaining  factors  of  the  various  terms. 


EXERCISES 

Factor  the  following : 

1.  6  Q?Wc  +  9  alPc^  -  15  a*6c7. 

Solution :  6  a'^hH  -t-  9  ah^c^  -  15  a^hc^  =  3  a6c  (2  at^  -J-  3  ftScS  -  5  a^<fi). 

2.  14  anx  —  21 6nic  —  7  n. 

3.  121  a'^hH  -  22  a^hc'^  +  H  oJb'^cK 

4.  5xV  -  10x8?/8  -  SxV  -  15x2y2. 

5.  21  ahn  +  6  obH"^  -  18  a'^hv?-  +  15  a'^h'^n. 

6.  10  al^cmx  -  5  ab^cy  +  6  ab'^cz  -  15  abc^m^. 
7  a^xV  -  49  ax^y*  +  14  axy^z^  -  21  a^'^y^. 

8.  45  a*62c8d  -  9a6*c«d8  -I-  27  a^bc*d^  -  117  a^l^cd*. 

16 


FACTORING  17 

30.  Factoring  by  grouping  terms.  If  in  the  expression  for 
the  distributive  law,     ac  +  bc  =  {a-\-  b)  c, 

we  replace  chj  c  -\-  d, 

we  have       a(G  -\-  d)  -\-  h  (c  -\-  d)  =  ac  -\-  ad  -\-  be  -\-  bd. 

We  may  then  factor  the  right-hand  member  as  follows  : 
ac_  -^  ad  -\-  be  -\-  bd  =  a  (c  -\-  d)  -\-  b  {c  -\-  d)  =  {a  -{-  b)  (c  -\-  d). 

This  affords  the 

EuLE.  Factor  out  any  monomial  expression  that  is  common  to 
each  term  of  the  polynomial. 

Arrange  the  terms  of  the  polynomial  to  be  factored  in  groups 
of  two  or  more  terms  each,  such  that  in  each  group  a  monomial 
factor  may  be  taken  outside  a  parenthesis  which  in  each  case 
contains  the  same  expression. 

Write  the  algebraic  sum  of  the  monomial  factors  that  occur 
outside  the  various  parentheses  for  one  factor,  and  the  expres- 
sion inside  the  parentheses  for  the  other  factor. 

EXERCISES 

Factor  the  following : 

1.  4a3&-6a262_4a4  +  6a63. 

Solution :  4  a^b  -  6  a'^h'^  -  4  a*  +  6  aft* 

=  2  a  (2  a26  -  3  a62  _  2  a3  +  3  63) 

=  2  a  (2  a26  -  2  a^  -  3  a62  +  3  &») 
=  2  a  [2  a2  (6  ^a)  +  3  62  (6  _  a)] 
=  2  a  (6  -  a)  (2  a2  +  3  62). 

2.  x2-(3a  +  46)x  +  12a6. 

Solution :  x2  -  (3  a  +  4  6)  x  +  12  a6 

=  x2  -  3  ax  -  4  6x  +  12  a6 
=  X  (x  —  3  a)  —  4  6  (x  —  3  a) 
=  (x-46)(x-3a). 

3.  2  ax  -  3  6y  -  2  6x  +  3  ay.  4.  56  a2  -  40  a6  +  63  ac  -  45  6c. 

5.  a''x2  —  6'?y3  _  f/jx^  +  apy\  6.  91  x2  -  112  mx  +  65  nx  -  80  mru 

7.  ax  -  bx  -[■  ex  +  ay  -  by  -\-  cy.  8.  2  ax  -  6x  -  c6  +  2  a6  +  2  ac  -  62. 


18  ALGEBRA  TO  QUADRATICS 

9.  2x6  -  3x8  +  2x«  -  3.  10.  x^  +  x^  +  x  +  1. 

11.  acx"^  -  hex  +  adx-bd.  12.  2  6x  -  x^  -  4  &  +  2  x. 

13.  3x3  _  12x82/2  _  4y2  +  1.  14.  4x2  -  (8  +  b)x  +  2  6. 

15.  x*  -  (4m  +  9n)x2  +  36 mn.  16.  x*  -  (2m  +  3n)x2  +  6mn. 

17.  a2x-ac+a6y-a62x-63y +c62.      18.  18  a»  -  2  ac*&6  _  g  a26  +  c^b^. 
19.  2ax  — 6ay  +  a-26x  +  56y-6.     20.  2ax-ay-26x  +  4cx— 2cy+62/. 

21.  2  a2x«  +  4  a2x*  +  2  a2x2  +  4  a^. 

22.  8x2  -  2 ox  -  12 xz  +  3az  +  4xy  -  ay. 

31.  Factors  of  a  quadratic  trinomial.  In  this  case  we  cannot 
factor  by  grouping  terms  immediately,  as  that  method  is  inappli- 
cable to  a  polynomial  of  less  than  four  terms.  We  observe,  how- 
ever, that  in  the  product  of  two  binomial  expressions. 

(mx  +  n)  (px  +  qy=  tnp^J  -h  {mq  +  w^)ic  -|-  ngy 

the  coefficient  of  x  is  the  sum  of  two  expressions  mq  and  np^ 

whose  product  is  equal  to  the  product  of  the  coefficient  of  x^  and 

the  last  term,  that  is, 

mq  •  np  =  m,p  •  nq. 

Thus,  to  factor  the  right-hand  member  of  this  equation,  we  may 
remove  the  parenthesis  from  the  term  in  x  and  use  the  principle 
rf  grouping  terms.    Thus 

mpx^  -f  (mq  +  np)  q?  +  nq 

=  TYipx^  -j-  m.qx  +  npx  -f-  nq 
=  mx  (px  -\-  q)+  n  (px  -f  q) 
=  (mx  +  n)  (px  -\-  q). 
This  affords  the 

Rule.  Write  the  trinomial  in  order  of  descending  powers  of 
X  (or  the  letter  in  which  the  expression  is  quadratic). 

Multiply  the  coefficient  of  a^  by  the  term  not  involving  x,  arid 
find  two  factors  of  this  product  whose  algebraic  sum  is  the 
coefficient  of  x. 

Replace  the  coefficient  of  x  by  this  sum  and  factor  by  group- 
ing terms. 


FACTORING  19 

Factoring  a  perfect  square  is  evidently  a  special  case  under  this  method. 
Thus  factor  x2  +  6x  +  9. 

1-9=9.  3  +  3  =  6. 

x2  +  (3  +  3)x  +  9 
=  x2  +  3x  +  3x  +  9 
=  a;(x  + 3)  +  3(x  +  3) 
=  (X  +  §)  (X  +  3) 

=  (X  +  3)2. 

One  will  usually  recognize  when  a  trinomial  is  a  perfect  square,  in  which 
case  the  factors  may  be  written  down  by  inspection. 


+  2  =  8. 


EXERCISES 

Factor  the  following : 

( 

1.  3x2  +  8x  +  4. 

/ 

Solution : 

3-4  = 

12.            /                   6  H 

3x2  +  8x  +  4 
=  3x2  +  (6  +  2)x  +  4 
=  3x2  +  6x  +  2x  +  4 
=  3x(x  +  2)  +  2(x  +  2) 
=  (3x  +  2)(x  +  2). 

2.  8x2- 

146X  +  352. 

Solution : 

8.  362:..  2462. 

8x2-146x  +  362 

26 -.126  =  -146. 

=  8x2-(2  6  +  12  6)x  +  362 
=  8x2-126x-26x  +  362 
=  4x(2x-36)-6(2x-36) 
=  (4x-6)(2x-36). 
3.  28x2 -3x- 40. 

Solution :  28  •  (-  40)  =  -  1120. 

The  factors  of  1120  must  be  factors  of  28  and  40.  We  seek  two  factors 
of  1120,  one  of  which  exceeds  the  other  by  3.  We  note  that  since  40  exceeds 
28  by  more  than  3,  one  factor  must  be  greater  and  the  other  less  than  28  and 
40  respectively. 

Since  4  •  7  =  28  and  5  ■  8  =  40, 

we  try  5  •  7=  35  and  4  •  8  =  82, 

which  are  the  required  factors  of  1120. 

28x2 -3x- 40 
=  28x2-(35-32)x-40 
=  28  x«  -  35  X  +  32  X  -  40 
=  7x(4x-6)  +  8(4x-6) 
=  (7x  +  8)(4x-5). 


20  ALGEBRA  TO  QUADRATICS 

4.  x2-6x  +  9.  5.  2x2  + re -6. 

6.  2x2  -  X  -  6.  7.  2x2  +  X  -  91. 

8.  x2  +  X  -  182.  9.  9x2  -  2x  -  7. 

10.  2x2  +  5a;  +  3.  11.  x2  -  llx  +  18. 

12.  3x2-10x-8.  13.  6x2  + 17 x  + 7. 

14.  16x2 +  4x- 3.  15.  7?/2-4?/-ll. 

16.  a2-6a6  +  962.  17.  5x2-12x  +  4. 

18.  18x2-73x  +  4.  19.  27x2 +  3x- 2. 

20.  24x2  -  31x  -  16.  21.  21x2  -  31x  +  4. 

22.  12  x2  +  60  X  -  72.  23.  x*  -  3  az^-A  2  a^. 

24.  9x2-18ax-7a2.  25.  10x2-63x-13. 

26.  x4y«  -  12 x2?/3  +  36.  27.  4x2p-16xi'- 81.                        >v 

28.  2 x8  -  17 6x2  +  sif'^x.  29.  4 a2  +  12 a6  +  9 62. 

30.  6  a2x2  -  2  a6x  -  7  62.    '  3 1.  10  x*  -  16  a2x3  -  100  x^a^. 

32.  4  aa"*  +  16  a^b»  +  16  62«.  33.  4  a^x^y*  -  20  a6xy2z  +  25  62z2. 

32.  Factoring  the  difference  of  squares.  Under  the  method  of 
the  preceding  paragraph  we  may  factor  the  difference  of  squares. 

Thus  to  factor  x^  —  b^  we  observe  that  the  product  of  the 
coefficient  of  x^  and  the  constant  term  is 

1 .  (_  b^)  =  _  b^. 

Since  the  coefficient  of  x.  is  zero,  we  have 

-b-\-b  =  0. 

Hence  (x  -\-  b)(x  —  b)—  x^  —  b\ 

EuLE.    Extract  the  square  root  of  each  term. 
The  sum  of  these  square  roots  is  one  factor,  and  their  differ- 
ence is  the  other. 

Example.  Factor  9a2xV  -  16  68c2. 

9  a2x«2/*  -  16  68c2  =  (3  axV  +  4  64c)  (3  ax-3y2  _  4  54c) . 

33.  Reduction  to  the  difference  of  squares.  The  preceding 
method  may  be  used  when  the  expression  to  be  factored  becomes 
a  perfect  square  by  the  addition  of  the  square  of  some  expression. 


FACTORING  21 

EXERCISES 

Factor  the  following : 

1.44  ^4  6*. 

Solution  r  a*  +  4  6*  =  a^  +  4  0,262  +  4  54  _  4  ^^252 

=  (a2  +  2b2)2_4a262 
=  (a2  +  2  62  _  2  ab)  {a^'-\-  2  62  +  2  db). 

2.  1  -  a*.  -  3.  a*  +  4. 

4.  x^  —  X.  5.  x«y4  +  4  x2. 

6.  4a;4+.2/*.  7.  4a2-2562. 

8.  x*  +  ic2  ^.  1.  9     ig  ^254  _  3.4. 

10.  x4  +  9x2  +  81.  11.  4a2p  -  962c2«. 

12.  a2p  +  3  -  16a36*.  13.  ar*«  +  x2«  +  1. 

14.  36x2^4^8  -  49  w2ui6.  15.  x4  -  13x2  +  36. 

16.  nty^  X IG  mx*  -  lij  m'x^y'^.  1 7.   9 x*  +  8  x^^  +  4  y^. 

34.  Replacing  a  parenthesis  by  a  letter.  Any  of  the  preceding  methods 
may  be  applied  when  a  polynomial  appeare  in  place  of  a  letter  in  the  expres- 
sion to  be  factored.  It  is  frequently  desirable  for  simplicity  to  replace  such 
a  polynomial  by  a  letter,  and  in  the  final  result  to  restore  the  polynomial. 

EXERCISES 

Factor  the  following : 

1.  2  ax2  -  2  6x2  -  6  ax  +  6  6x  -  8  a  +  8  6. 

Solution :  2  ax2  -  2  6x2  -  6  ax  +  6  6x  -  8  a  +  8  6 

=  2(a  — 6)x2  -  6(a  -  6)x  -  8(a  -  6) 
=  (a -6)  (2x2 -6.x -8) 
=  2{a-6)(x2-3x-4) 
=  2  (a  -  6)  (X  -  4)  (X  +  1). 

In  this  example  the  factor  (a  —  6)  might  have  been  replaced  by  a  letter. 

2.  a2  +  62  -  c2  -  9  -  2  a6  +  6  c. 

Solution :  a2  +  62  -  c2  -  9  -  2 a6  +  6c 

=  a2  -  2 a6  +  62  -  (c?  -  6c  +  .9) 
=  (a  -  6)2  -^  (c  -  3)2  . 

=  (a  -  6  +  c  -  3)  (a  -  6  -  c  +  g. 

3.  (3x-2/)(2a+p)-(3x-2/)(a-9). 

4.  (4a-66)(3m-2p)  +  (a  +  56)(3m-2p). 

5.  (7a-32/)(5c-2d)-(6a-22/)(5c-2d). 

6.  (X  -  2/)  (3a  +  46)  -  (4a  -  56)  (X  -  2/)  -  (X  -  y)  (2a  -  86). 


22  ALGEBRA  TO  QUADRATICS 

7.  6(x  +  2/)2-ll(x4-2/)-7. 

8.  4a2  -  12a6  +  962  _  x2  -  2x  -  1. 

9.  x2a2  +  2  x2a  +  x2  -  a2  -  2  a  -  1. 
10.  ax2  +  6 ox  +  9a  -  &x2  -  66x  -  96. 

fll.  4  (a  -  6)2  -  5(a2  -  62)  -  21  (a  +  6)2. 

12.  6(x  +  ?/)2  -  12  (x2  -y2)^4{x-  yf. 

13.  a262x2  -  a262  -  2  a6x2  +  2  a6  +  x2  -  1. 

14.  (x-2y)(2a-36)-(96-10)(x-2?/). 

/     35.  Factoring  binomials  of  the  form  d^  ±  6*^.    By  §  27, 

b/ ^^     V        ci"  +  ^"  =  («  +  ^'Xa"-^  -  a"-2^  +  a«-3^»2 _|_  ^,«-i>)^ 

where  w  is  odd. 

One  can  factor  by  inspection  any  binomial  of  the  given  form 
by  reference  to  these  equations. 

EXERCISES 

Factor  the  following : 

1.  x^  —  2/6. 

Solution :  x^  -  ye  -  ^xB  _  ^3)  (a;3  +  ^3) 

=  (X2  +  Xy  +  7/2)  (X  -  y)  (X2  -  Xy  +  2/2)  (x  _|.  y). 

2.  x6  +  125.  3.  x^  -  1. 
4.  xi2  -  2/12.  5.  x^  —  2/9. 

6.     Xl8   -  2/18.  7.     a;16   _  yl6. 

8.  a2x3  +  a5.  9.  x*  -  a82/4. 

.  10.  216  a  +  a*.  11.  ox*  -  16  a. 

12.  3a7  -  96  65a2. 1^-""  13.  2,1  x^^  +  64  2/8. ^"^^ 

14.  27  x52/7  +  x22/4.  «--^  15.  16a468-81ci6d8,/^ 

36.  Highest  common  factor.  An  expression  that  is  not  further 
divisible  into  factors  with  rational  coefficients  is  called  prime. 

If  two  polynomials  have  the  same  expression  as  a  factor,  this 
expression  is  said  to  be  their  common  factor. 

The  product  of  the  common  prime  factors  of  two  polynomials 
is  called  their  highest  common  factor,  or  H.C.F. 

The  same  common  prime  factor  may  occur  more  than  once.  Thus  (x  —  1)  2  (x + 1) 
and  (x  -  1)2  (a;  -  2)2  have  (x  -X^  as  their  H.C.F.  v-^— 


c^f 


FACTORING  23 

37.  H.C.F.  of  two  polynomials.  The  process  of  finding  the 
H.C.F.  is  performed  as  follows: 

EuLE.  Factor  the  polynomials.  The  product  of  the  common 
prime  factors  is  their  H.C.F. 

EXERCISES  [l^'^-Xi^\)i^^^UH 

Find  the  H.C.F.  of  the  following: 

1.  4  ab^x^  -  8  ab^x^  +  4  ab^  and  6  abx"^  +  12  abx  +  6ab. 
Solution :  —  4  ab'^x^  -  8  ab^^  +  4  ab^ 

=  4a62(x4- 2x2+1) 
=  4a62(x-l)2(x  +  l)2. 

6  a6x2  +  12  abx  +  6ab 
=  6a6(x2  +  2x  +  1) 
=  6a6(x  + 1)2. 

The  H.C.F.  is  then  2  a6(x  +  1)2. 

2.  x^  —  y^  and  x2  —  y"^. 

3.  x3  +  x2  -  12 X  and  x2  +  5x  +  4. 

4.  9mx2  —  Qmx  +  m  and  9nx2  —  n. 

5.  6x- 4x2  + 2 ax -3a  and  9  -  4x2. 

6.  12  a2  -  3(3a6  +  27  62  and  8a2  -  1862. 

7.  3  a2x  -  6  a6x  +  3  62x  and  4  02^/  -  4  62y. 

8.  2  X  -  46  -  x2  -  2  6x  and  4x  -  5x2  -  6. 

9.  6x^-7  ax2  -  20  a2x  and  3 x2  +  ax  -  4  a2. 

38.  Euclid's  method  of  finding  the  H.C.F.  When  one  is  unable  to  factor 
the  polynomials  whose  H.C.F.  is  sought,  the  problem  may  nevertheless  be 
solved  by  use  of  a  method  which  in  essence  dates  from  Euclid  (300  e.g.). 

The  validity  of  this  process  depends  on  the  following 
Principle.   If  a  polynomial  has  a  certain  factor,  any  multiple  of  it  has  the 
same  factor. 

Let  x»  +  ^x»-i +  5x»-2  +  . . . +£: 

and  .  x«  + ax'»-i  +6x'»-2  H 1- ^ 

be  represented  by  F  and  6?  respectively.    The  letters  A,  B,    ■  ■,  K  and  a,  6, 

•  •  •,  Z  represent  integers,  and  m,  the  degree  of  G,  is  no  greater  than  n,  the 

degree  of  F.    We  seek  a  method  of  finding  the  H.C.F.  of  F  and  G  if  any 

exists.    Call  Q  the  quotient  obtained  by  dividing  F  by  G,  and  call  B  the 

remainder.     Then  (§  26) 

^'  F=QG-^B,  (1) 


24  ALGEBRA  TO  QUADRATICS 

where  the  degree  of  E  in  x  is  not  so  great  as  that  of  G.  Now  whatever  the 
H.C.F.  of  i^'and  G  may  be,  it  must  also  be  the  H.C.F.  of  G  and  B.    For  since 

F-QG  =  B, 

the  H.C.F.  of  F  and  G  must  be  a  factor  of  the  left-hand  member,  and  hence 
a  factor  of  R,  which  is  equal  to  that  member.  Also  every  factor  common 
to  G  and  R  must  be  contained  in  F,  for  any  factor  of  G  and  E  is  a  factor  of 
the  right-hand  member  of  (1),  and  hence  of  F. 

Thus  our  problem  is  reduced  to  finding  the  H.  C.  F.  of  G  and  R.  Let  Qi  and 
Bi  be  respectively  the  quotient  and  remainder  obtained  in  dividing  G  by  R. 

Then  (?  =  QiB  4-  Ru 

where  the  degree  of  Ei  in  x  is  not  as  great  as  that  of  B.  By  reasonirg  simi- 
lar to  that  just  employed  we  see  that  the  H.C.F.  of  G  and  R  is  also  the 
H.C.F.  of  B  and  Bi.     Continue  this  process  of  division. 

Let  B  =  Q2R1  +  R2, 

-Bi  =  QsRi  +  Eg. 
until,  say  in  Rk  =  Qk+2Rk  +  \  +  Bk+i, 

either  B^  is  exactly  divisible  by  B^  +1  (i.e.  E^-  +  2  =  0),  or  i?;t  +  2  does  not  con- 
tain X.  This  alternative  must  arise  since  the  degrees  in  x  of  the  successive 
remainders  E,  Ei,  E2,  •  •  •  are  continually  diminishing,  and  hence  either  the 
remainder  must  finally  vanish  or  cease  to  contain  x.  Suppose  Bk  +  2=0.  Then 
the  H.C.F.  of  Bk  and  Bk  + 1  is  Ea.-  + 1  itself,  which  must,  by  the  reasoning  given 
above,  be  also  the  H.C.F.  of  F  and  G.  If  Rk  +  2  does  not  contain  x,  then  the 
H.C.F.  of  F  and  G,  which  must  also  be  a  factor  of  Ea  +  2,  can  contain  no  x, 
and  must  therefore  be  a  constant. 

Thus  F  and  G  have  no  common  factor  involving  x. 

This  process  is  valid  if  the  coefficients  of  F  and  G  are  rational  expressions  in 
any  letters  other  than  x. 

39.  Method  of  finding  the  H.C.F.  of  two  polynomials.  The  above  dis- 
cussion we  may  express  in  the  following 

Rule.  Divide  the  polynomial  of  higher  degree  {if  the  degrees  of  the  polyno- 
mials are  unequal)  by  the  other,  and  if  there  is  a  remainder,  divide  the  divisor 
by  it ;  if  there  is  a  remainder  in  this  process,  divide  the  previous  remainder 
by  it,  and  so  on  until  either  there  is  no  remainder  or  it  does  not  contain  the  letter 
of  arrangement.  If  tJiere  is  no  remainder  in  the  last  division,  the  last  divisor  is 
the  H.  C.  F.  If  the  last  remainder  does  not  contain  the  letter  of  arrangement, 
then  the  polynomials  have  no  common  factor  involving  that  letter. 

In  the  application  of  this  rule  any  divisor  or  remainder  may  be  multiplied  or 
divided  by  any  expression  not  involving  the  letter  of  arrangement  without  affect- 
ing the  H.C.F. 


FACTORING 


?6 


EXERCISES 
Find  the  H.C.F.  of  the  following: 

2x 


1.  2x*  +  2x3-x2 
Solution :  /Q  ^ 

Multiply  by  - 


1  andx4  +  x3  +  4x  +  4. 

2J2x*  +  2x3-x2-   2x-l|xH     x^+  4x  +     4 

2x^  +  2x3  +   8x  +  8 

■1,  -x2-10x-9    *■  ^'       ^ 

x^-\-10x+9\x^+     x3+  4x  +     4|x2-9x  +  81 


Ri 


x4  +  10x3+  9x2 


,-  [^-«) 


-  9x3-  9x2+     4x  ^t4H/l 

-  9x3-^90x2-  81x  I 


Divide  by  -  725, 


(9-!.     c ,      ec, 

x  +  9|x2  +  lOx  +  9|x  +  l  \ 


81x2+  85X+4 
81x2  +  810x+729 
-726ig    7a6 


X2  + 


>y 


/'te.F. 


9x  +  9 

9x  +  9 

0        0 
Thus  the  H.C.F.  is  x  +  1. 
This  process  may  be  performed  in  the  following  more  compact  form. 


2 

2x4  +  2x3-x2-    2x-l 

a^+      x^+     4x  +     4 

x2-9x  +  81 

2x*  +  2x3           +    8x  +  8 
-x2  -  lOx-9 

X4  +  I0x3+        9X2 

-1 

-    9x^-      9x2+      4x 

x  +  9 

x2  +  10x  +  9 

-   9x3-    90x2-    81 X 

X2+         X 

81x2+    85x  +      4 

9x+9 
9x  +  9 

81x2+810x  +729 

-725x   -725 

-725 

0         0 

X  +1 

Result:   x  +  1. 

2.  x2  +  6 X  -  7  and  x^  -  39x  +  70. 

3.  x^  —  X*  —  X  +  1  and  5  x*  —  4  x^  —  1. 

4.  x8  +  2x2  +  9  and  -  6x3  -  iia;2  +  i5x  +  9. 

5.  x3  -  2x2  -  15x  +  36  and  3x2  -  4x  -  15. 

6.  x*  -  3x3  +  x2  +  3x  -  2  and  4 x3  -  9x2  +  2 X  +  3. 

7.  4x8  -  18x2  +  19x  -  3  and  2x*  -  12x3  +  19x2  _  6x  +  9. 

8.  X*  +  4x3  _  22x2  -  4x  +  21  and  x*  +  10x8  +  20x2  -  lOx  -  21. 

9.  6  a*x3  -  9  a3x2y  -  10  a^  xy^  +  16  ay^  and  10  a^xV  -  1^  «*«^y*  +  8  a^*y^ 
12  a2x3y6. 


26  ALGEBRA  TO  QUADRATICS 

40.  Least  common  multiple.    The  least  common  multiple  of  two 

or  more  polynomials  is  the  polynomial  of  least  degree  that  con- 
tains them  as  factors.  We  may  find  the  least  common  multiple 
of  several  polynomials  by  the  following 

EuLE.  Multiply  together  all  the  factors  of  the  various  poly- 
nomials, giving  to  each  factor  the  greatest  exponent  with  which  it 
appears  in  any  of  the  polynomials, 

41.  Second  rule  for  finding  the  least  common  multiple.  When 
only  two  polynomials  are  considered  the  previous  rule  is  evidently 
equivalent  to  the  following 

EuLE.  Multiply  the  polynomials  together  and  divide  the 
product  hy  their  highest  common  factor. 

EXERCISES 

Find  the  least  common  multiple  of  the  following : 

1.  x2  -  y2,a;2  +  ay  —  ax  -  xy,  and  a;2  -  2  xy  +  y\ 

Solution :  x^ -y'^  =  {x  -  y)ix -^  y). 

x'^  +  ay  -ax-xy  =  {x-  y)  {x  —  a). 
x^-2xy  +  y^  =  {x-  yf. 

Thus  the  L.C.M.  =  (x  -  yY  (x  +  y)  (x  -  a). 

2.  4a26c,  6a62,  and  12  c2 

3.  9x?/2,  6x22/3,  and  ZxyH"^. 

4.  (X  +  1)  (x2  -  1)  and  x3  -  1. 

5.  x*  +  4x22/2  and  x2  +  2 2/2  -  2 y. 

6.  4x2  -  9?/  and  4x2  -  12x2/  +  9?/2. 

7.  x2  -  4  X  +  3,  x2  -  1,  and  x2  -  ax  —  x  +  a. 

8.  x-1,  2x2-5x-3,  and2x8-7x2  +  2x  +  3. 

9.  x*  -  9x2  +  26x  -  24  and  x^  -  10x2  +  31  x  -  30. 

10.  2x2  -  3x  -  9,  x2  -  6x  +  9,  and  3x2  _  9x  -  6x  +  3&. 


CHAPTEE  III 
FRACTIONS 

42.  General  principles.  The  symbolic  statements  of  the  rules 
for  the  addition,  subtraction,  multiplication,  and  division  of  alge- 
braic fractions  are  the  same  as  the  statements  of  the  correspond- 
ing operations  on  numerical  fractions  given  in  (2),  (3),  and  (4), 
§  6.  This  is  immediately  evident  if  we  keep  in  mind  the  fact 
that  algebraic  expressions  are  symbols  for  numbers  and  that  if  the 
letters  are  replaced  by  numbers,  the  algebraic  fraction  becomes  a 
nunierical  fraction. 

43.  Principle  I.  Both  numerator  and  denominator  of  a  frac- 
tion may  he  multiplied  (or  divided)  hy  the  same  expression  with- 
out changing  the  value  of  the  fraction. 

This  follows  from  (5),  §  6. 

44.  Principle  II.  If  the  signs  of  both  numerator  and  denomi- 
nator of  a  fraction  he  changed,  the  sign  of  the  fraction  remains 
unchanged. 

This  follows  from  Principle  I,  when  we  multiply  both  numerator  and 
denominator  by  —  1 . 

45.  Principle  III.  If  the  sign  of  either  numerator  or  denomi- 
nator (hut  not  hoth)  he  changed,  the  sign  of  the  fraction  is  changed. 

This  follows  from  (6),  §  6. 

46.  Reduction.  A  fraction  is  said  to  be  reduced  to  its  lowest 
terms  when  its  numerator  and  denominator  have  no  common 
factor.    We  effect  this  reduction  by  the  following 

Eule.  Divide  hoth  numerator  and  denominator  hy  their 
highest  common  factor. 

27 


28  ALGEBRA  TO  QUADRATICS 

EXERCISES 
*    Reduce  the  following  to  their  lowest  terms. 
12  ax2  -  12  ab^ 


1. 


4ax^-Sabx-\-4:al» 

12ax2-12a62  12 a {x  -  b) {x  ■{■  b) 


4  ax2  -  8  a6x  +  4  ab^ 


Solution : 


H.C.F.  =4a{x-b). 

a;S  +  4x        *^  a  —  2ax  —  lOx  +  5 

g    6x2  -8ax  +2a2  2  a26  +  2  a62  -  2  abc 

x2-a2  *  ■   3  6c2  -  3  62c  -  3  abc  ' 

g    a2  +  62  _  c2  +  2  a&  21x8-9x2  +  7x  -  3 

xi8  -  ai8*  •  a2  -  62  4.  c2  +  2 ac'        '    3x8  +  15x2  +  x  +  5  * 

^j^    2x2+  3x-9     j^2      X*  -  x8  -  X  +  1        j^„    x3  +  3ax2  +  3a2x  +  a8 
x2-9       '        '  2x4 -x3-2x  +  l'        *  a2  +  2ax  +  x2 

47.  Least  common  denominator  of  several  fractions.  We  have 
the  following 

Rule.  Find  the  least  common  multiple  of  the  various  denomi- 
nators. 

Multiply  both  numerator  and  denominator  of  each  fraction  hy 
the  expression  which  will  make  the  new  denominator  the  least 
common  multiple  of  the  denominators. 


EXERCISES 

Reduce  the  following  to  their  least  common  denominator. 

-    2         8  ^   2x-3 

1.  ->  ,  and  — -. 

X    2x-l  4x2-1 

Solution  :  The  L.C.M.  of  the  denominators  is  x (4x2  —  1) .    Thus  the  frac- 
tions are 

2 (4 x2  -  1)     3x(2x+l)       ^^  x(2x-3) 

x(4x2-l)'   x(4x2-l)  '  ^^     x(4x2-l)' 
"  ,  and 3.  ,  ■ ,  and 


6  +  a  a2-62  2x-8    4x2  +  4x-16  4x2-26 


FRACTIONS  29 


A        1  1  ^        1 

4.  ,   ,  and 


aj3  _  y3    aj4  _  y4  a;2  -  y2 


5.  ,  — - — ,  and 


c        2x-\            X  ^    2x-3 

6.  — ,  ,  and 


x2  -  2  X  +  1    x2  -  1  (x  +  1)2 

-a                  b  .  c 

7.  : ,  ,  and 


a  +  b-c    a-\-b  -^c  d^  +  2ab  +  b^  -  c"^ 

48.  Addition  of  fractions.  This  operation  we  perform  as 
follows : 

EuLE.  Reduce  the  fractions  to  he  added  to  their  least  common 
denominator. 

Add  the  numerators  for  the  numerator  of  the  sum,  and  take 
the  least  common  denominator  for  its  denominator. 

49.  Subtraction  of  fractions.  This  operation  we  perform  as 
follows : 

Rule.  Reduce  the  fractions  to  their  least  common  denomi- 
nator. 

Subtract  the  numerator  of  the  subtrahend  from  that  of  the 
minuend  for  the  numerator  of  the  result,  and  take  the  least 
common  denominator  for  its  denominator. 

50.  Multiplication  of  fractions.  This  operation  we  perform 
as  follows : 

Rule.  Multiply  the  numerators  together  for  the  numerator 
of  the  product,  and  the  denominators  for  its  denominator. 

51.  Division  of  fractions.  This  operation  we  perform  as 
follows : 

Rule.  Invert  the  terms  of  the  divisor  and  multiply  by  the 
dividend.  a 

Remark.    Since  a  fraction  is  a  means  of  indicating  division,  t~^^  and  - 

are  two  expressions  for  the  same  thing. 

d 


30  ALGEBRA  TO  QUADRATICS 


EXERCISES  ^ 

Perform  the  indicated  operations  and  bring  the  results  into  their  simplest 
forms. 

a  +  b     a  —  b 


I    I*  —  w      a  +  b 

'  a  +  b      a  —  b 


a  +  & 


Solution ; 


a  4-  6  a  —  b 
a  ^b  a  +  6 
a  +  b  a  —  b 
a—b     a+b 

a^-l^ 


{a  +  6)2  +  (a  ■ 

-6)2 

a2-62 

(a  +  6)2  _  (a  . 

-6)2 

a2-62 

(a  +  6)2  +  (a  - 

-6)2 

a2-62 

( 

2  a2  +  2  62      a2  +  62 

4a6 

2a6 

iW  - 1 

2.V    * 

3  +  f  +  i 

(a  +  6)2  -  (a  -  6)2 


'  1  +  1                          '     2.V    *  •  1-|-t\ 

5       2-V-  g     3  +  f  +  l  7    2  +  f  +  f 

*  i-(-W                 ■f  +  l  +  i.V  '1-l  +  f" 

8.  ^_^L±^.  9.  ^-i^-^.  10.  -i-  + 


6        26  ab      ac      be  a  —  b      a  +  b 

11.1+1+1.      12. '?^l^+!-^      13.  20= 


a6c  c            c  a-la2-l 

-,    3a;-l      2«-7  --  2ic-l      2x-5 

14. .  15. 

l-3x          7  a;-2a;-4 

16.—?^ L_.  17.  ^                8 


4x-4      6a;  +  6  3x-9      6«-15 

18    «  +  ^  ■  g'^  +  2 a6  - 62  2a-36      3a-26 

'  a-6  ■        a2-62       "  "      12 a     "*"     16a 

20    ?. ■         Q  21      «t?  +  ^c  acZ-6c 


15(x-l)      10(x  +  l)  2cd{c-d)      2cd{c  +  d) 

22    a?  +  y     a;-y        4zy  ^^    a;2  +  a;(a  +  6)  +  a6  gg-gg 
'  x-y      x-\-y      x2-y2'  '  x2  -  x(a  +  6)  +  a6' x^  -  62' 

24.  i5  +  il  +  i?£-l^.  25.  ^  « 


6  a      14  a      36  a      16  a  a2  -  9  a  +  14  '  a*  -  6  a  -  14 

26.  Z^  +  lU/^-l  +  lV  27        g(«-'g)  a(a  +  x) 

\y8     a;/     \y2     y     x)  a2  +  2 ttx  +  x2   a2-2ax  +  xa* 

28.  lzi^.lz^./i  +  _±_V  29.  -A_  +  _^  +  _A._l. 
1  +  6    a  +  a«   V       1-a/  (x  - 1)«      (x  - 1)2  ^  x  - 1      x 


FRACTIONS  31 


^  1 

30.  a  +  --L..  31.  -^^i 32.  ^. 

c  +  ^  (^Y-1  x  +  l  +  l 


e                                 \h/  X 

33.  1 34. 35.  a  + ^ 

1                                    .1  d 

x-\ 4h cH 

x-S  9 

a      ,      h      ^  db                  1,1  a2  +  62    ^ 


o«a  +  6      a-6      a2-62  i  +  x      1-x     „„        a  a2-62 

OO.  •    Of.  •     So.  ' 

1  1  1111       a3  +  63 


(a  +  &)2      {a-hY  1  - «      1  +  ic  a      h 

39     •     a;-3y        ^        x  +  3y  ^^    _J^ a;  +  2 

■  x2-2a;y-15?/2  "  x2-8xi/  +  15y2'         '  3(x  +  l)      3(-4-3x  +  x2) 

^^'   1 ri'+         2y.       ;• 

X     y  +  z 
-Q    2a-36  +  4      3a-46  +  5      a-1 

6  8  12 

43    l^"^  -^y^     x2  -  y2\      /x  +  y     X  -  y\ 
■  \x2  -  2/2      a;2  4. 2/2/  •  \^a;  _  y      x^yj' 

X- 1      y- 1      z -1 

-^  3xy2 x__         y g 

'  yz  -{•  zx  —  XV  11      1 

X     y     « 
45    /2x  +  y      2y-x  x2     \      x2  +  y2 

\  X  +  y        X  -  y       x2  -  y2/  ■  x2  -  y2 
-^a  —  36      4a  —  6      5a  +  3c      a2  —  6c      2a 

40. 1 1 • 

6a  26  9c  2ac  6 

.  „  6cd!  cda 

47. 1 ^ 

(a  -  6)  (a  -  c)  (a  -  d)      (6  -  c)  (6  -d^(b-a) 

dab  ^ 


(c-6)(c-a)(c-d) 
a6c 
"^  {d  -a){d-  6)  (d  -  c) ' 


48     1       a-26       8       3a-4c2       9        5c2_--_66 
'  6a        3a6        46  8ac2  8c2         126c2 


'    X-1         X  +  1         (X  -  1)2         (X  +  1)2        X2  -  1         (X2  -  1)2 


CHAPTER  IV 


EQUATIONS 

52.  Introduction.  An  equation  is  a  statement  of  equality 
between  two  expressions. 

We  assume  the  following 

Axiom.  If  equals  he  added  tOy  subtracted  frorriy  multiplied  hy^ 
or  divided  hy  equals,  the  results  are  equal. 

As  always,  we  exclude  division  by  zero.  In  dividing  an  equation  by  an  alge- 
braic expression  one  must  always  note  for  what  values  of  the  letters  the  divisor 
vanishes  and  exclude  those  values  from  the  discussion. 

53.  Identities  and  equations  of  condition.  Equations  are  of 
two  kinds  : 

First.  Equations  that  may  be  reduced  to  the  equation  1  =  1  by 
performing  the  indicated  operations  are  called  identities. 

Thus  2  =  2, 

a -6=  (3a -26)  -(2a -6) 

are  equations  of  this  type.  In  identities  the  sign  =  is  often  replaced  by  =.  It 
should  be  noted  that  identities  are  true  whatever  numerical  values  the  letters 
may  have. 

Second.  Equations  that  cannot  be  reduced  to  the  form  1  =  1, 
but  which  are  true  only  when  some  of  the  letters  have  particular 
values,  are  called  equations  of  condition  or  simply  equations. 

Thus  z=2  cannot  further  be  simplified,  and  is  true  only  when x  has  the  value 
2.  Also  a;  =  2  a  is  true  only  when  x  has  the  value  2  a  or  a  has  the  value  -  •  If  in 
this  equation  x  is  replaced  by  2  a,  the  equation  of  condition  reduces  to  an  identity. 

The  number  or  expression  which  on  being  substituted  for  a 
letter  in  an  equation  reduces  it  to  an  identity  is  said  to  satisfy 
the  equation. 

Thus  the  number  5  satisfies  the  equation  x^  —  24  =  1.  The  number  3  satisfies 
the  equation  (x  —  3)  (a;  +  4)  =  0. 

32 


EQUATIONS  33 

The  process  of  finding  values  that  satisfy  an  equation  is  called 
solving  the  equation.  The  development  of  methods  for  the  solu- 
tions of  the  various  forms  of  equations  is  the  most  important 
question  that  algebra  considers. 

In  an  equation  in  which  there  are  two  letters  it  may  be  possible 
to  find  a  value  which  substituted  for  either  will  satisfy  the  equa- 
tion.   Thus  the  equation  x  —  2  a  =  Ois  satisfied  if  x  is  replaced  by 

2 -a,  or  if  a  is  replaced  by  -•   In  the  former  case •  we  have  solved 

for  £c,  that  is,  have  found  a  value  that  substituted  for  x  satisfies 
the  equation.  In  the  latter  case  we  have  solved  for  a.  In  any 
equation  it  is  necessary  to  know  which  letter  we  seek  to  replace 
by  a  value  that  will  satisfy  the  equation,  that  is,  with  respect  to 
which  letter  we  shall  solve  the  equation. 

The  letter  with  respect  to  which  we  solve  an  equation  is  called 
the  variable. 

Values  which  substituted  for  the  variable  satisfy  the  equation 
are  called  roots  or  solutions  of  the  equation. 

When  only  one  letter,  i.e.  the  variable,  occurs  in  an  equation,  the  root  is  a  num- 
ber. When  letters  other  than  the  variable  occur,  the  root  is  expressed  in  terms 
of  those  letters. 

54.  Linear  equations  in  one  variable.  An  equation  in  which 
the  variable  occurs  only  to  the  first  degree  is  called  a  linear  equa- 
tion. To  solve  a  linear  equation  in  one  variable  we  apply  the 
following 

EuLE.  Apply  the  axiom  (§  52)  ^o  obtain  an  equation  in  which 
the  variable  is  alone  on  the  left-hand  side  of  the  equation. 

The  right-hand  side  is  the  desired  solution. 

To  test  the  accuracy  of  the  work  substitute  the  solution  in  the 
original  equation  and  reduce  to  the  identity  1=1. 

Since  the  result  of  adding  two  numbers  is  a  definite  number,  and  the  same  is 
true  for  the  other  operations  used  in  finding  the  solution  of  a  linear  equation,  it 
appears  that  every  linear  equation  in  one  variable  has  one  and  only  one  root. 

When  both  sides  of  an  equation  have  a  common  denominator,  the  numerators 
are  equal  to  each  other.  This  appears  from  multiplying  both  sides  of  the  equation 
by  the  common  denominator  and  then  canceling  it  from  both  fractions. 


34  ALGEBRA  TO  QUADRATICS 

EXERCISES 

Solve : 

-    4«-2  ,  6a;     Sx      ^ 

^•-^  +  T  =  T  +  '- 

Solution :  Transpose  the  term  involving  x, 

4a;-2      5x      3x      ^ 

...  .      ^.  32x-16  +  25x-30a;      ^ 

Add  fractions,     =  5. 

40 

Clear  of  fractions  and  simplify,        97  x  —  216 

x  =  8. 
rt    a(d^  +  x^)  ,  ax 

dx  d 

d^  4-  x^  X 

Solution :  Divide  by  a,  =  c  +  -. 

dx  d 

Transpose  the  term  involving  x, 

d*  +  x2       X 


dx         d 


=  c. 


Add  fractions,  =  c. 

dx 

Clear  of  fractions  and  simplify,  ,  _ 

d 

X  =  -' 

c 

3.  {a-l)x  =  b-x.  4.  {a-x){l-x)  =  z^-l. 

5.  a(x-a2)  =  6(x-62).  6.  2a;  -  fx  =  f «  -  1  -  |x  +  2. 

7.  8x  -  7  +  X  =  9x  -  3  -  4x.  8.  .617 x  -  .617  =  12.34  -  1.234x. 

9.  3(2x-.3)  =  .6  +  5(x-.l).  10.  7  -  5x  +  10  +  8x  -  7  +  3x  =  x. 
11.  (x-3)(x-4)  =  (x-6)(x-2).  12.  f {x\[|(|x+5)-10]  +  3}-8  =  0. 
13.  (H-6x)2+(2  +  8x)2  =  (H-10x)2.   14.  6  =  3x+i(x  +  3)-^(llx-37). 

15.  2(x  +  5) (X  +  2)  =  (2x  +  7)(x  +  3). 

16.  (7ix  -  2|)  -  [4|  -UH-  5a;)]  =  18^. 

17.  6x  -  7(11  -  X)  +  11  =  4x- 3(20 -X). 

18.  (a  -  6)  (X  -  c)  +  (a  +  6)  («  +  c)  =  2(&x  +  ad). 

19.  2x  -  3(6  +  f  X)  +  ^(4  -  X)  -  ^(3x  -  16)  =  0. 

20.  5x-2  =  fx  +  fx  +  fx  +  T'ffa;  +  H«  +  i|a;. 

21.  (a  -  6)  (a  -  c  +  x)  +  (a  +  &)  (a  +  c  -  X)  =  2  a*. 

22.  12.9x  -  1.46X  -  3.29  -  .99x  -  llx  +  .32  =  0. 

23.  6.7x  -  2^7.8  -  9.3x)  =  5.38  -  4|(.28  +  3.6x). 


EQUATIONS  S6 

24.3-^  =  ^-.  25.  -^+l^c. 

3      11  mx      nx 


3      X 
26 


x-J:^x-^^  27    iil^^lll^?. 

x-S      x-^'  '  f (6x  +  l)      3' 

28     t(a^-4)   ^1  29    2x2-3x  +  5^2 

'1(3x4-5)      6*  ■7x2-4x-2      7* 

3o.»  +  l  =  12  +  l  31.  i:i-%i  =  ^--l 

x2x9  J+x4^+x4 

„rt«  +  &aj_c  +  cZx  ««25x  — 2_52x  — 5 

a  +  b  ~  c  +  d'  '  3"7x-3~  7    3x-  7* 

„-    ox      ex  ,  /x      ,  „-    X  +  a      &      X  -  6      a 

34.  —  +  —  -^  —  =  h.  35.  — = +  -■ . 

bag  b         a         a         b 

36.  5^  +  i,  =  x-l.  3y    3x-19^5x-25^3 

a  X  —  13         X  +  7 


3      12      1         3      2 


38.  L4-L^  =  -l--L.        39.  ^—^  +  ^-^i^  +  '-:z^  =  o. 


3      12      12    1 

2'^x      3'^x      S'x"^ 


^^x-8x  +  12      „,      18  ,-a(2x  +  l)      Sox- 46      4 

4U. 1- =  z  H •  41.    —  =  — • 

x  +  2       x-8  x  +  2  36  66  6 


.„    ax      6x       2a6       (a  +  6)2x 

'4(6. i f-  =  < 

6        a       a  +  6  ab 

43.  8ix  -  -  -  3|x-  4^x  +  1  =  0. 

6 

..        1         a  +  b         1         a  —  b 

44. + = + 

a  +  b         x         a  —  b         x 

45    ^ (^  -  ^)  _L  ^  +  8  _  3(5x  +  16) 
X-  7        X  -4~     5X-28   ' 

46.^ 


a  6 

^^^ox-A_^ 


X      b^c  —  X      «c2  —  X  _ 


5X-.4      1.3 -3x_  1.8-8X 


2  1.2 

^48.  -^^  + -1^- + -^  =  2. 
6x  +  2      15x4-6      3x4-1 

49.I^^2-i(.  +  3)  +  6  =  ^-(^±?>. 
3  6^  '  2 


36 


ALGEBRA  TO  QUADRATICS 


50. 
51. 
52. 
53. 
54. 
55. 
56. 
57. 
58. 
59. 
60. 
61. 
62. 
63. 
64. 
65. 
66. 
67. 


5x-l        3x  +  2       ic2-30x  +  2 


3(x  +  l)      2(x-l)  6x2-6 

3x-2  .  7x-3  .      x  +  100 


=  10. 


x  +  3  x  +  2       x2  +  5x  +  6 
36(x-a)      x-62      6(4a  +  cx)^Q 

5a  166              6a 

16X-27  x  +  3_6  +  3x      4x-7 

21  6     ~       2                3      " 

a{b-x)  b{c-x)  _  a  +  6  ,  /&  ,  a\ 

6x  ex              X         \c      &/ 

5x-6  9-lOx      3x-4      3-4x 


10 

4-2x 

3 

3- 


6 
1.5x 


7 
4x2 


6x-3      X-.5      3(2x-l) 


^  + 


6(2x-5)      2(2x-6)  3(2x-6) 

x«  +  i-3x»-i      3x«-i-x«  x«      „ 

4x                        4  2 

ax  —  be      6x  —  ac_cx  —  62  x  —  a            x 

ab              c^              be  c                a 


3x 


X  — 2a  X  — 26  X  — 2c 


6  +  c 


a+c— 6      a+6— c      a+6+c 


2x«  +  7x»-i      7x"-44x«-i      4X"  +  27x'»-i 


9  5X-14 

3x  +  3      /x  +  1 


18 


a(x-3)      6(x-3)      a2(x-l)      62(x-l)^^ 
6  a  62        "^       a2        ~    • 


(m  +  n)2x      nx  _        c  3nx 

mb  6       m  (a  —  6)        6 


m  (a  —  6)  6 

4(13x-.6)      3(1.2-x)_9x  +  .2      5  +  7x 

6  "^  2  -~20~"^~~r~"^'^- 

(^^±^(x-a)  +  ^^i:iA'(x-6)  =  2a(2a  +  6-x). 
6  a 

ax  —  6       ex  — d      (6n  +  dm)x  +  (6p  +  dg)  _  a      e 
mx—p      nx  —  q  (mx  -  p)  (nx  -  g)       ~"  m      n' 


EQUATIONS  37 

55.  Solution  of  problems.  The  essential  step  in  solving  a 
problem  by  algebra  is  the  expression  of  the  conditions  of  the 
problem  by  algebraic  symbols.  This  is,  in  fact,  nothing  else  than 
a  translation  of  the  problem  from  the  English  language  into  the 
language  of  algebra.  The  translation  should  be  made  as  close  as 
possible,  clause  by  clause  in  most  cases.  In  general  the  result 
sought  should  be  represented  by  the  variable,  which  for  that 
reason  is  often  called  the  unknown  quantity. 

Example.  What  number  is  it  whose  third  part  exceeds  its  fourth  part  by 
sixteen  ? 

Solution :  "  What  number  is  it "  is  translated  by  x.  Thus  we  let  x  represent 
the  number  sought.    "  Whose  third  part "  is  translated  by  - .    "  Exceeds  its 

XX 

fourth  part "  is  translated  by ,  i.e.  the  third  part  less  the  fourth  part 

leaves  something.   "  By  sixteen  "  gives  us  the  amount  of  the  remainder.  Thus 
the  translation  of  the  problem  into  algebraic  language  is 
Let  X  represent  the  number  sought. 

^-^  =  16. 
3      4 

This  equation  should  be  solved  and  checked  by  the  methods  already  given. 


PROBLEMS 

1.  What  number  is  it  whose  third  and  fifth  parts  together  make  88  ? 

2.  What  number  increased  by  3  times  itself  and  5  times  itself  gives  99  ? 

3.  What  is  the  number  whose  third,  fourth,  sixth,  and  eighth  parts 
together  are  3  less  than  the  number  itself? 

4.  What  number  is  it  whose  double  is  7  more  than  its  fourth  part  ? 

5.  In  10  years  a  young  man  will  be  3|  times  as  old  as  his  brother  is 
now.    The  brother  is  7|  years  old.    How  old  is  the  young  man  ? 

6.  A  father  who  is  53  years  old  is  3  years  more  than  12 1  times  as  old  as 
his  son.    How  old  is  the  son  ? 

7.  If  you  can  tell  how  many  apples  I  have  in  my  basket,  you  may  have 
4  more  than  ^,  or,  what  is  the  same  thing,  4  less  than  i  of  them.  How  many 
have  I? 

8.  If  Mr.  A  received  ^  more  salary  than  at  present,  he  would  receive 
$2100.    How  much  does  he  receive? 

9.  A  boy  spends  ^  of  his  money  in  one  store  and  \  of  what  remains  in 
another,  and  has  24  cents  left.    How  much  had  he  ? 


38  ALGEBRA  TO  QUADRATICS 

10.  A  man  who  is  3  months  past  his  fifty-fifth  birthday  is  4^  times  as  old 
as  his  son.    How  old  is  the  son  ? 

11.  In  a  school  are  four  classes.  In  the  first  is  ^  of  all  the  pupils ;  in  the 
second,  ^ ;  in  the  third,  j- ;  in  the  fourth,  37.  How  many  pupils  are  in  the 
school ? 

12.  A  merchant  sold  to  successive  customers  ^,  |,  and  ^  of  the  original 
length  of  a  piece  of  cloth.  He  had  left  2  yards  less  than  half.  How  long 
was  the  piece  ? 

13.  How  may  one  divide  77  into  two  parts  of  which  one  is  2|  times  as 
great  as  the  other  ? 

14.  The  sum  of  two  numbers  is  73  and  their  difference  is  15.  What  are 
the  numbers  ? 

15.  A  father  is  4i  times  as  old  as  his  son.  Father  and  son  together 
are  27  years  younger  than  the  grandfather,  who  is  71  years  old.  How  old  are 
father  and  son  ? 

16.  The  sum  of  two  numbers  is  999.  If  one  divides  the  first  by  9  and 
the  second  by  6,  the  sum  of  these  quotients  is  138.    What  are  the  numbers  ? 

17.  The  first  of  two  numbers  whose  sum  is  a  is  b  times  the  second. 
What  are  the  numbers  ? 

18.  If  the  city  of  A  had  14,400  more  inhabitants,  it  would  have  3  times 
as  many  as  the  city  of  B.  Both  A  and  B  have  together  12,800  more  than 
the  city  of  C,  where  there  are  172,800  inhabitants.  How  many  are  in 
A  and  B  ? 

19.  Two  men  who  are  26  miles  apart  walk  toward  each  other  at  the 
rates  of  3|  and  4  miles  an  hour  respectively.    After  how  long  do  they  meet  ? 

20.  A  courier  leaves  a  town  riding  at  the  rate  of  6  miles  an  hour.  Seven 
hours  later  a  second  courier  follows  him  at  the  rate  of  10  miles  an  hour. 
How  soon  is  the  first  overtaken  ? 

21.  A  can  copy  14  sheets  of  manuscript  a  day.  When  he  had  been  work- 
ing 6  days,  B  began,  copying  18  sheets  daily.  How  many  sheets  had  each 
written  when  B  had  finished  as  many  as  A  ? 

22.  The  pendulum  of  a  clock  swings  387  times  in  5  minutes,  while  that 
of  a  second  clock  swings  341  times  in  3  minutes.  After  how  long  will  the 
second  have  swung  1632  times  more  than  the  first? 

23.  The  difference  in  the  squares  of  two  numbers  is  221.  Their  sum  is 
17.    What  are  the  numbers  ? 

24.  If  a  book  had  236  more  pages  it  would  have  as  many  over  400  pages 
as  it  now  lacks  of  that  number.    How  many  pages  has  the  book  ? 

25.  A  man  is  now  63  years  old  and  his  son  21.  When  was  the  father 
19  times  as  old  as  his  son  ? 


EQUATIONS  39 

'  26.  If  7  oranges  cost  as  much  less  than  50  cents  as  13  do  more  than 
50 cents,  how  much  do  they  cost  apiece? 

27..  The  numerator  of  a  fraction  is  6  less  than  the  denominator.  Dimin- 
ish both  numerator  and  denominator  by  1  and  the  fraction  equals  |.  Find 
the  fraction. 

28.  The  sum  of  three  numbers  is  100.  The  first  and  second  are  respec- 
tively 9  and  7  greater  than  the  third.    What  are  the  numbers  ? 

29.  Out  of  19  people  there  were  f  as  many  children  as  women,  and  1^ 
times  as  many  men  as  women.    How  many  were  there  of  each  ? 

30.  A  boy  has  twice  as  many  brothers  as  sisters.  His  sister  has  5  times 
as  many  brothers  as  sisters.    How  many  sons  and  daughters  were  there  ? 

ySl.  A   dealer  has  5000  gallons    of    alcohol    which    is  85%    pure.  .  He 
/wishes  to  add  water  so  that  it  will  be  75%  pure.    How  much  water  must 
he  add?  i 

V  32.  How  much  water  must  be  added  to  5  quarts  of  acid  which  is  10%  full 
strength  to  make  the  mixture  8|%  full  strength  ? 

.^33.  A  merchant  estimated  that  his  supply  of  coffee  would  last  12  weeks. 
He  sold  on  the  average  18  pounds  a  week  more  than  he  expected,  and  it 
lasted  him  10  weeks.    How  much  did  he  have  ? 

34.  At  what  time  between  3  and  4  o'clock  are  the  hands  of  a  clock  point- 
ing in  the  same  direction  ? 

35.  At  what  time  between  11  and  12  o'clock  is  the  minute  hand  at  right 
angles  to  the  hour  hand  ? 

^^6.  A  merchant  bought  cloth  for  |2  a  yard,  which  he  was  obliged  to  sell 
for  $1.75  a  yard.  Since  the  piece  contained  3  yards  more  than  he  expected, 
he  lost  only  2%.    How  many  yards  actually  in  the  piece  ? 

37.  A  man  has  three  casks.  If  he  fills  the  second  out  of  the  first,  the 
latter  is  still  f  full.  If  he  fills  the  third  out  of  the  second,  the  latter  is  still 
I  full.  The  second  and  third  together  hold  100  quarts  less  than  the  first. 
How  much  does  each  hold  ? 

38.  A  crew  that  can  cover  4  miles  in  20  minutes  if  the  water  is  still,  can 
row  a  mile  downstream  in  f  the  time  that  it  can  row  the  mile  upstream. 
How  rapid  is  the  stream  ? 

39.  A  cask  is  emptied  by  three  taps,  the  first  of  which  could  empty  it 
in  20  minutes,  the  second  in  30  minutes,  the  third  in  35  minutes.  How  long 
is  required  for  all  three  to  empty  the  cask  ? 

40.  A  can  dig  a  trench  in  f  the  time  that  B  can  ;  B  can  dig  it  in  f  the 
time  that  C  can ;  and  A  and  C  can  dig  it  in  8  days.  How  long  is  required 
by  all  working  together  ? 


40  ALGEBRA  TO  QUADRATICS 

56.  Linear  equations  in  two  variables.  A  simple  equation  in 
one  variable  has  one  and  only  one  solution,  as  we  have  abeady 
seen  (p.  33).  On  the  other  hand,  an  equation  of  the  first  degree 
in  two  variables  has  many  solutions. 

For  example,  Sx  +  7y  =  l 

is  satisfied  by  innumerable  pairs  of  numbers  which  may  be  sub- 
stituted for  X  and  y.    For,  transposing  the  term  in  y,  we  get 

X-      3     , 

from  which  it  appears  that  when  y  has  any  particular  numerical 
value  the  equation  becomes  a  linear  equation  in  x  alone,  and 
hence  has  a  solution.  Thus,  when  y  =  1,  ic  =  —  2,  and  this  pair 
of  values  is  a  solution  of  the  equation.  Similarly,  cc  =  —  9,  y  =  4 
also  satisfy  the  equation. 

57.  Solution  of  a  pair  of  equations.  If  in  solving  the  equation 
just  considered,  the  values  of  x  and  y  that  one  may  use  are  no 
longer  unrestricted  in  range,  but  must  also  satisfy  a  second  linear 
equation,  we  get  usually  only  a  single  pair  of  solutions.  Thus 
if  we  seek  a  solution,  that  is,  a  pair  of  values  of  x  and  y  satisfying 

Sx-\-7y  =  l, 

such  that  also 

x  +  y  =  -ly 

we  find  that  the  pair  of  values  x  =  —  2,  y  =  1  satisfy  both  equa- 
tions. Any  other  solution  of  the  first  equation,  as,  for  instance, 
x=—9,  y  =  4,  does  not  obey  the  condition  imposed  by  the  second. 
Two  equations  which  are  not  reducible  to  the  same  form  are 
called  independent. 

Thus  6a; -82/ -4  =  0 

and  3  a;  —  4  y  =  2 

are  not  independent,  since  the  first  is  readily  reduced  to  the  second  by  transposing 

and  dividing  by  2.   They  are,  in  fact,  essentially  the  same  equation.    On  the  other 

hand,  .         „ 

'  X  —  4y=2 

and  3  X  —  4  y  =  2 

are  not  reducible  to  the  same  form  and  are  independent.  Since  dependent  equa- 
tions are  identical  except  for  the  arrangement  of  terms  and  some  constant  factor, 
all  their  solutions  are  common  to  each  other. 


=  -% 


EQUATIONS  41 

This  principle  we  may  state  as  follows : 

Two  equations  ,  i,     ,         r. 

ax  +  oy  +  c  =  0 

and  a'%  +  &'?/  +  C  =  0 

are  dependent  when  and  only  when 

£_  &^  _  c 

Independent  equations  in  more  than  one  variable  which  have 
a  common  solution  are  called  simultaneous  equations. 

Two  pairs  of  simultaneous  equations  which  are  satisfied  by  the 
same  pair  (or  pairs)  of  values  of  x  and  y  and  only  these  are  called 
equivalent. 

Thus  r3.+7a==l.     ^^      ra==- 

are  equivalent  pairs  of  equations. 

58.  Independent  equations.   We  now  prove  the  following 

Theorem.    If  A=0  and  B  =  0  represent    two    in 
equations^  then  the  pairs  of  equations 

^  =  ^'(1)     and     {'''■^f^'!:  (2) 

are  equivalent  where  a,  b,  c,  and  d  are  any  numbers  such  that 
ad  —  be  is  not  equal  to  zero. 

The  letters  A  and  B  symbolize  linear  expressions  in  x  and  y. 
Evidently  any  pair  of  values  of  x  and  y  that  makes  both  A  =  0  and 
B  =  0,  i.e.  satisfies  (1), also  makes  aA-{-bB=0  and  cA  +  dB  =  0, 
i.e.  also  satisfies  (2).  We  must  also  show  that  any  values  of  x 
and  y  that  satisfy  (2)  also  satisfy  (1). 

For  a  certain  pair  of  values  of  x  and  y  let 

aA-{-hB  =  0,  (3) 

cA  +  dB=  0.  (4) 

Multiply  (3)  by  c  and  (4)  by  a  (§  52). 

Then  acA  +  bcB  =  0,  (5) 

acA  +  adB  =  0.  (6) 


42  ALGEBRA  TO  QUADRATICS 

Subtract  (5)  from  (6)  (§  52), 

(ad-bc)B  =  0. 
Thus,  by  §  5,  either  ad  —  be  =  0     ot     B  =  0. 

But  ad  —  be  is  not  zero,  by  hypothesis ;  consequently  -6  =  0. 
Similarly  we  could  show  that  ^  =  0. 

Thus  if  we  seek  the  solution  of  a  pair  of  equations  ^  =  0, 
jB  =  0,  we  may  obtain  by  use  of  this  theorem  a  pair  of  equiva- 
lent equations  whose  solution  is  evident,  and  find  immediately 
the  solution  of  the  original  equations. 

59.  Solution  of  a  pair  of  simultaneous  linear  equations.   The 

foregoing  theorem  affords  the  following 

KuLE.  Multijply  each  of  the  equations  by  some  number  such 
that  the  coejfficunts  of  one  of  the  variables  in  the  resulting  pair 
of  equations  are  identical. 

Subtract  one  equation  from  the  other  and  solve  the  resulting 
simple  equation  in  one  variable. 

Find  the  value  of  the  other  variable  by  substituting  the  value 
just  found  in  one  of  the  original  equations. 

Check  the  result  by  substituting  the  values  found  for  both 
variables  in  the  other  equation. 

Example. 

Solve  3x-l-7y  =  l,  (1) 

x  +  y=-\.  (2) 

Solution :  Multiply  (1)  by  1  and  (2)  by  3, 

3x  +  7?/  =  l, 
3x  +  32/  =  -3 


Subtract, 

4y  = 

4 

Substitute 

in  (2), 

x  +  l  = 

1. 
-1. 

Check:  Substitute 

in  (1), 

«  = 

-2. 

3. 

(- 

-  2)  +  7  . 1  = 

-6  +  7 

60.  Incompatible  equations.   Equations  in  more  than  one  vari- 
able that  do  not  have  any  common  solution  are  called  incompatible. 


EQUATIONS  43 

Theokem.    The  equations 

ax-\-by  =  c,  ^  (1) 

afx-\-b'i/  =  c'  '*(2) 
are  incompatible  when  and  only  when  aV  —  ba^  —  0. 

Apply  the  rule  of  §  59  to  find  the  solution  of  these  equations. 
Multiply  (1)  by  a'  and  (2)  by  a. 

We  obtain  aa^x  +  a^by  —  ca\ 

aa'x  -f-  ab'y  =  ac'. 
Subtract,  (ab'  —  a'b)  y  =  ac'  —  ca'. 

If  now  ab'  —  a'b  is  not  zero,  we  get  a  value  of  y ;  but  since  under 
our  hypothesis  ab'  —  a'b  =  0,  we  can  get  no  value  for  y  since  divi- 
sion by  zero  is  ruled  out  (§  7).    Thus  no  solution  of  (1)  and  (2) 
exists. 
Example. 

Solve  Sx  +  1y  =  l,  (1) 

6x  +  14y  =  l.  (2) 

Solution :   Multiply  (1)  by  2, 

Qx  +  Uy  =  2 
6x+  Uy=  1 
Subtract,  0=1 

which  is  absurd.    Thus  no  solution  exists. 

61.  Resum^.    We  observe  that  pairs  of  equations  of  the  form 
ax  -\-  by  -{-  c  =  Oj 
a'x  -f  J'y  +  c'  =  0 
fall  into  three  classes  : 

(a)  Dependent   equations,  which   have   innumerable    common 
solutions.  ^       ^       ^ 

Tlien  a'^b'^7''  W 

(b)  Incompatible  equations,  which  have  no  common  solution. 
Then  ^^,  _  ^,j  ^  ^^  ^^^  ^^^  -^  ^^^  ^^^^^ 

(c)  Simultaneous  equations,  which  have  one  and  only  one  pair 
of  solutions. 

Then  ab'  -  a'b  ^  0. 


18 


44  ALGEBRA  TO  QUADRATICS 

EXERCISES 

Solve  and  check  the  following : 
.    2x  +  6y  =  l,  2    4x-6y,=  8,  ,    6x  +  8y 

'  Qx  +  1y  =  S.  '  |x-2/  =  f.  •x  +  |y  =  3. 

^    7x-3y  =  27,  ^    2x-|2/  =  4,  ^    |y  =  ^x-l, 

*  6x-6y  =  0.  ■3x-|2/=:0.  '^-^yrzfx-l, 

„    5x-4y  +  l  =  0,  g    3x  +  4y  =  253,         g    6x  +  3y  +  2  =  0, 

1.7x-2.22/  +  7.9  =  0.       'y  =  5x.  ■3x  +  2y  +  l=0. 

j^Q    x  +  my  =  a,  j^j^    x  +  y  =  |(5a  +  6), 

'  x  —  ny  =  b.  '  X  —  y  =  |(a  +  6 6). 

-2    2x-3y  =  -5a,  -,    |x-i(y  +  l)  =  l, 

•  3x-22/  =  -5&.  •  i(a;  +  l)  +  |(y-l)  =  9. 

--    3.5x  +  2i2/=13  +  4fx-3.5y,       --    3x  +  2  ?/ =  5a2  +  a6  +  5&2^ 
'  2ix  +  .8y  =  22^  +  .7x-3i2/.  "  3?/ +  2x  =  6a2  -  a6  +  662. 


16. 

3      8 

x+r 

16  _  4 
X       2/ 

=  3, 
=  4. 

Hint.  Ketain  fractions. 

18. 

X  -  c 
y-C 

a 
6' 

r 

x-y  = 

a  - 

h. 

n/\ 

x  +  2y 
2x-2/ 

+  1 
+  1 

=  2, 

17. 


1     1_6 

-  4-  -  —  -1 
X      y      6 

111 

X      y      Q 


3x  +  l_4 


19.  4-2y      3 
X  +  y  =  1. 

-  +  ^  =  c, 

20.  r"^"^;  21.  "^      ^ 

^^-^  +  ^  =  5.  ■        ^  +  ^  =  ci. 

X  —  2/  +  3  «!      6i 

5  7 


22. 


24. 


x  +  2y      2x  +  2/ 

7^6 
3x-2~6-2/* 
.9x-.7y  +  7.3_ 
13X-152/  +  17" 
1.2x-.2y  +  8.9^  ^ 
13x-16y  +  17 
X  y  1 


.2. 


a  +  6     a  —  6     a  —  6 

X     _     y     _     1 
a  +  6      a  —  6      a  +  6 


x+1      a+6+c 

2.3 

y  +  l~a-6  +  c' 

X  —  1      a  +  6  —  c 

y-1      a-6— c 

25. 

y    _6 

27. 

a- 0     a  - c 
a2  -  X     a^-y  _ 

EQUATIONS  46 

^^^    y  =  4  -  3x  +  x2.  *  •  (4a;  -  7) (X  -  3)  =  y. 

„Q4Vx-3Vy  =  6,  g-xVa-y\/6  =  a  +  6, 

3  Vx  —  4  Vy  =  1.  ,      X  -\-y  =  2  Va. 

22    xV2  +  ?/V3  =  3V3,  32    4V^T7-5Vir^  =  7, 

'xV3-y\/2  =  2V2.  "3  VxTT  -  7  Vy^  =  2. 

«^    Vx      Vy  „>    Vx  -  3      Vy  +  3 

12,  ^^4  9  , 

—=  +  —=  =  1.  ,  +     ,  =  4. 

Vx      Vy  Vx  -  3      V  2/  +  3 

.        ,,     ,  a  +  6  +  1  x  +  ly  +  2      2(x-y) 

(a_&)[x  +  (a+&)y]=a-6+l.  3(x  -  3)-4(y- 3)  =  12(2y -x> 

62.  Solutions  of  problems  involving  two  unknowns.  Tlie  same 
principle  of  translation  of  the  problem  into  algebraic  symbols 
should  be  followed  here  as  in  the  solution  of  problems  leading  to 
simple  equations  (p.  37). 

PROBLEMS 

1.  The  difference  between  two  numbers  is  3|.  Their  sum  is  9|.  What 
are  the  numbers? 

2.  What  are  the  numbers  whose  sum  is  a  and  whose  difference  is  6  ? 

3.  A  man  bought  a  pig  and  a  cow  for  $100.  If  he  had  given  $10  more  for 
the  pig  and  $20  less  for  the  cow,  they  would  have  cost  him  equal  amounts. 
What  did  he  pay  for  each  ? 

4.  Two  baskets  contain  apples.  There  are  51  more  in  the  first  basket  than 
in  the  second.  But  if  there  were  3  times  as  many  in  the  first  and  7  times  as 
many  in  the  second,  there  would  be  only  6  more  in  the  first  than  in  the 
second.    How  many  apples  are  there  in  each  basket? 

5.  A  says  to  B,  "Give  me  $49  and  we  shall  then  have  equal  amounts." 
B  replied,  "If  you  give  me  $49,  I  shall  have  3  times  as  much  as  you. 
How  much  had  each? 

6.  A  man  had  a  silver  and  a  gold  watch  and  two  chains,  the  value  of  the 
chains  being  $9  and  $25.  The  gold  watch  and  the  better  chain  are  together 
twice  and  a  half  as  valuable  as  the  silver  watch  and  cheaper  chain.  The 
gold  watch  and  cheaper  chain  are  worth  $2  more  than  the  silver  watch  and 
the  better  chain.    What  is  the  value  of  each  watch  ? 


46  ALGEBRA  TO  QUADRATICS 

7.  What  fraction  is  changed  into  ^  when  both  numerator  and  denomi- 
nator are  diminished  by  7,  and  into  its  reciprocal  when  the  numerator  is 
increased  by  12  and  the.  denominator  decreased  by  12  ? 

8.  A  man  bought  2  carriage  horses  and  5  work  horses,  paying  in  all 
$1200.  If  he  had  paid  $5  more  for  each  work  horse,  a  carriage  horse 
would  have  been  only  J  more  expensive  than  a  work  horse.  How  much 
did  each  cost? 

9.  A  man's  money  at  interest  yields  him  $540  yearly.  If  he  had  received 
\%  more  interest,  he  would  have  had  $60  more  income.  How  much  money 
has  he  at  interest  ? 

10.  A  man  has  two  sums  of  money  at  interest,  one  at  4%,  the  other  at  5%. 
Together  they  yield  $750.  If  both  yielded  1%  more  interest,  he  would  have 
$165  more  income.    How  large  are  the  sums  of  money  ? 

11.  A  man  has  two  sums  of  money  at  interest,  the  first  at  4%,  the  second 
at  3^%.  The  first  yields  as  much  in  21  months  as  the  second  does  in  18 
months.  If  he  should  receive  \%  less  from  the  first  and  \%  more  from  the 
second,  he  would  receive  yearly  $7  more  interest  from  both  sums.  What 
are  the  sums  at  interest? 

12.  What  values  have  a  mark  and  a  ruble  in  our  money  if  38  rubles  are 
worth  14  cents  less  than  75  marks,  and  if  a  dollar  and  a  ruble  together 
make  %\  marks  ? 

13.  A  chemist  has  two  kinds  of  acid.  He  finds  that  23  parts  of  one  kind 
mixed  with  47  parts  of  the  other  give  an  acid  of  84 1%  strength  and  that 
43  parts  of  the  first  with  17  parts  of  the  second  give  an  80f%  pure  mixture. 
What  per  cent  pure  are  the  two  acids  ? 

14.  Two  cities  are  30  miles  apart.  If  A  leaves  one  city  2  hours  earlier 
than  B  leaves  the  other,  they  meet  1\  hours  after  B  starts.  Had  B  started 
2  hours  earlier,  they  would  have  met  3  hours  after  he  started.  How  many 
miles  per  hour  do  they  walk  ? 

15.  The  crown  of  Hiero  of  Syracuse,  which  was  part  gold  and  part  silver, 
weighed  20  pounds,  and  lost  1^  pounds  when  weighed  in  water.  How  much 
gold  and  how  much  silver  did  it  contain  if  19;^  pounds  of  gold  and  10^ 
pounds  of  silver  each  lose  one  pound  in  water? 

16.  Two  numbers  which  are  written  with  the  same  two  digits  differ  by 
36.  If  we  add  to  the  lesser  the  sum  of  its  tens  digit  and  4  times  its  units 
digit,  we  obtain  100.    What  are  the  numbers  ? 

17.  A  company  of  14  persons,  men  and  women,  spend  $48.  If  each  man 
spends  $4  and  each  woman  $3,  how  many  men  and  how  many  women  are  in 
the  company? 


EQUATIONS  47 

63.  Solution  of  linear  equations  in  several  variables.  This 
process  is  performed  as  follows : 

KuLE.  Eliminate  one  variable  from  the  equations  taken  in 
pairs,  thus  giving  a  system  of  one  less  equation  than  at  first 
in  one  less  variable. 

Continue  the  process  until  the  value  of  one  variable  is  found. 

The  remaining  variables  may  be  found  by  substitution. 

Special  cases  occur,  as  in  the  case  of  two  variables,  where  an  infinite  number 
of  solutions  or  no  solutions  exist.  Where  no  solution  exists  one  is  led  to  a  self- 
contradictory  equation  on  application  of  the  rule.    See  exercise  17,  p.  48. 

EXERCISES 

Solve  and  check  the  following : 

«  +  2/  +  2  =  9, 
1.  a;  +  2y +  42  =  16, 
x  +  3y +  9z  =  23. 

Solution :        x  +     y  -[■     z  =    Q  x  -{■     y  ■}-     z—    ^ 

x  +  2y  -{-iz  =  15  x  +  3y-f  9z=:23 

2/  +  32=    6  2y  +  8z  =  U 

y  +  4z=    7 

y  +  Sz  =  6 

y  -\-4z  =  7 

2=1 

y  +  3  =  6. 

y  =  3. 

a;  +  3  +  1  =  9. 

x  =  S. 

Check :   5  +  9  +  9  =  23. 

X  +  y  =  37,  x  +  y  =  xy, 

2.x +  2  =25,  3.  2x  +  2z=x2;, 

y  +  z  =  22.  Sz  +  Sy  =  zy. 

Hint.  Divide  the  equations  by  xy,  xz,  yz  respectively. 

X  +  y  4-  z  =  17,  X  +  y  +  z  =  36, 

4.  X  +  z  -  y  =  13,  5.  4x  =  3y, 

x  +  2;-2y  =  7.  2x  =  3z. 

L3x-1.9y=.l,  2x+2y  +  «  =  a, 

6.  1.7y-l.lz  =  .2,  7.  2y-i-2z+x  =  6, 

2.9x-2.1z  =  .3.  2z+2x  +  y  =  c. 


48  ALGEBRA  TO  QUADRATICS 


x  +  2y  =  6, 

8. 

y  +  22;  =8, 

z  +  2u  =  ll, 

u  +  2x  =  Q. 

y     z 

10. 

i  +  l  =  25, 

X      z 

1  +  1  =  20. 
X     y 

"y      _oo 

iy-Zx--''' 

12. 

.  ^'       -15, 

9. 


X  +  y  =  m, 
y  +  z  =  a, 
z  +  u  =  n, 
u  —  x  =  b. 


11. 


xy    _1 

;  +  y~5' 
xz        1 


x  +  z     6 

yz    _  1 

y-{-z~  1 

y  +  1 

13.?^  =  4, 
X-32  '  z  +  1 

yz       ^^2.  2;  +  3_l 


4y-5z  x  +  1      2 

14.  312/  =  X  +  2  +  12,  15.  X  +  2  =  2|  2/  -  14, 


4^  2  =  X  +  y  +  16.  y  +  2;  =  3f  X  -  32. 

x  +  2y-z=4.6,  x  +  2y  +  32  =  15, 

y  +  22;-x  =  10.1,  17.  3x  +  5y  +  7z  =  37, 
2  +  2x-y  =  6.7.  5x  +  By +  llz  =  59. 

7x  +  6y  +  72  =  100,  (x+2)(2y  +  l)  =  (2x+7)y, 

x_2y  +  2  =  0,  19.  (x-2)(3z  +  l)  =  (x+3)(32-] 
3x  +  y-2«  =  0.  (l+l)(«+2)  =  (y+3)(z+l). 


CHAPTEE  V 
RATIO  AND  PROPORTION 

64.  Ratio.    The  ratio  of  one  of  two  numbers  to  the  other  is  the 
result  of  dividing  one  of  them  by  the  other. 

a 
The  ratio  of  a  to  6  is  denoted  by  a :  6  or  by  -  • 

The  dividend  in  this  implied  division  is  called  the  antecedent, 
the  divisor  is  called  the  consequent. 

65.  Proportion.    Four  numbers,  a,  b,  c,  d,  are  in  proportion  when 
the  ratio  of  the  first  pair  equals  the  ratio  of  the  second  pair. 

This  is  denoted  by  a :  6  =  c :  d  or  by  -  =  -  • 

The  letters  a  and  d  are  called  the  extremes,  b  and  c  the  means, 
of  the  proportion. 

66.  Theorems  concerning  proportion.   If  a,  b,  c,  d  are  in  pro- 
portion, that  is,  if 

a:b  =  G:doT-  =  -j  (I) 

b      d  ^  ^ 

then  ad  =  be,  (II) 

b:a  =  d:c,  '          (III) 

a:c  =  b:d,  (IV) 

a  -\-  b  :  a  =  c  -{-  d  :  Cf  (V) 

a  —  b  :  a  =  c  —  d  :  Cj  (VI) 

a-{-b:a  —  b=:c-{-d:c  —  d:  C^II) 

Equation  (III)  is  said  to  be  derived  from  (I)  by  inversion. 
Equation  (IV)  is  said  to  be  derived  from  (I)  by  alternation. 
Equation  (V)  is  said  to  be  derived  from  (I)  by  composition. 
Equation  (VI)  is  said  to  be  derived  from  (I)  by  division. 
Equation  (VII)  is  said  to  be  derived  from  (I)  by  composition 
and  division. 

49 


50  ALGEBRA  TO  QUADRATICS 

67.  Theorem.  If  a  numher  of  ratios  are  equal,  the  sum  of  any 
number  of  antecedents  is  to  any  antecedent  as  the  sum  of  the 
corresponding  consequents  is  to  the  corresponding  consequent. 

Let  a\h  —  c\d  =  e\f=  g '.h^ 


or 

a      c       e      g 
'h~d~f~h' 

To  prove 

a  +  c  +  e      b-hdi-f 
g                h 

If 

a      c      e      g 
1~'d~f~h~''' 

we  have 

a  =  hr, 

c  =  dr, 

g  =  hr. 
Divide  the  sum  of  the  first  three  equations  by  the  last  and  we  get 
gj^c  +  e  ^b  +  d-\-f 
g  h 

68.  Mean  proportion.    The  mean  proportional  between  two  num- 
bers a  and  c  is  the  number  b,  such  that 

a:b  =^b  :  c. 
By  (II),  §  66,  we  see  that  ac  =  b^. 


EXERCISES 

If  a  :  6  =  c  :  d,  prove  that : 

a2  .    ,       c2 


1.  a  +  h: =  c  +  d: 

a  +  b 

Solution :  By  (V),  §  66, 
Squaring,  we  get 


a  +  b  c  +  d 

a  -hb  _c  +  d 
a  c 

(g  +  b)^  ^  (c  +  d)^ 
a2  c2      ' 


a  +  b  _c  +  d 
^^  a2     ~     c2    ' 


a  +  6      c  +  d 
a  +  b: -  =  c  +  d 


a  +  b  c-\-  d 


I 


RATIO  AND  PROPORTION  61 

2.  a2  :  62  =  c2  :  d2.  3.  a  +  b:c  +  d  =  a:c. 

4.  ma:mh  =  nc:  nd.  5.  a2 :  c2  =  a2  +  62  .  c2  +  d^. 

6.  a2  +  62 :  _^  =  c2  +  d52 :  _^.     7.  Va2  +  c^ :  VPTd^  =  a  :  6. 
a  +  6  c  +  cZ 

8.  ma  +  n6  :  ra  +  s6  =  mc  +  n(i  :  re  +  sd. 

9.  a  +  h  +  c  +  d:a  —  b-\-c  —  d  =  a  +  b  —  c  —  d:a  —  h  —  c  +  d. 

10.  Find  the  mean  proportional  between  a^  +  c^  and  62  +  ^2. 

11.  Find  the  mean  proportional  between  a2  +  62  +  c^  and  62  +  c2  +  d\ 

Solve  the  following  f or  x : 

12.  20:96  =  x:  57. 
14.  x  —  ax:Vx  =  Vx  :  x. 

-  „    Vx  +  7  +  Vx      4  +  Vx 

10.     — — ^;;^ = 

V^rp^  -Vx4-v^  a-6a6  ac 

Hint.  Use  composition  and  division. 

18.  (?^^^  +  „i)..^±^-„i  =  (a  +  bf:z. 
\a  —  0  I     a  +  6 


13. 

8  a6 :  X  : 

=  6c  :  1|  ac. 

15. 

\--/x 

l-3Vx  = 

=  1 

4. 

17. 

a  +  &   fl 

^2-62 
=  X  : 

a  - 

-6 

CHAPTER  VI 

IRRATIONAL  NUMBERS  AND  RADICALS 

69.  Existence  of  irrational  numbers.  We  have  seen  that  in 
order  to  solve  any  linear  equation  or  set  of  linear  equations  with 
rational  coefficients  we  need  to  make  use  only  of  the  operations 
of  addition,  subtraction,  multiplication,  and  division.  When, 
however,  we  attempt  to  solve  the  equation  of  the  second  degree, 
x^  =  2,  we  find  that  there  is  no  rational  number  that  satisfies  it. 

Assumption.  A  factor  of  one  memher  of  an  identity  between 
integers  is  also  a  factor  of  the  other  memher. 

Thus  let  2  .  a  =  6,  where  a  and  b  are  integers.  Then  since  2  is  a  factor  of  the 
left-hand  member,  it  must  also  be  contained  in  6. 


Theorem.    iVb  rational  number  satisfies  the  equation  x^  =  ^. 
Suppose  the  rational  number  -  be  a  fraction  reduced  to  i' 
lowest  terms  which  satisfies  the  equation.    Then 


or  a2  =  2  b\  (1) 

Thus,  by  the  assumption,  2  is  contained  in  a^j  and  hence  in  a. 
Suppose  a  =  2  a'. 

Then  by  (1)  4  a'^  =  2  b% 

or  2  a'^  =  h\ 

that  is,  2  must  also  be  contained  in  ft,  which  contradicts  the 

hypothesis  that  t  is  a  fraction  reduced  to  its  lowest  terms. 

The  fact  that  the  equation  x^  =  2  has  no  rational  solution  is 
analogous  to  the  geometrical  fact  that  the  hypotenuse  of  an 
isosceles  right  triangle  is  incommensurable  with  a  leg. 

62 


IRRATIONAL  NUMBERS  AND  RADICALS  63 

70.  The  practical  necessity  for  irrational  numbers.  For  tlie 
practical  purposes  of  the  draughtsman,  the  surveyor,  or  the 
machinist,  the  introduction  of  this  irrational  number  is  superflu- 
ous, as  no  measuring  rule  can  be  made  exact  enough  to  distin- 
guish between  a  length  represented  by  a  rational  number  and  one 
that  cannot  be  so  represented.  As  the  draughtsman  does  not  use 
a  mathematically  perfect  triangle,  but  one  of  rubber  or  wood,  it 
is  impossible  to  see  in  the  fact  of  geometrical  incommensurability 
just  noted  a  practical  demand  from  everyday  life  for  the  intro- 
duction of  the  irrational  number.  In  fact  the  irrational  number 
is  a  mathematical  necessity,  not  a  necessity  for  the  laboratory  or 
draughting  room,  as  are  the  fraction  and  the  negative  number. 
We  need  irrational  numbers  because  we  cannot  solve  all  quad- 
ratic equations  without  them,  and  the  practical  utility  of  those 
nimibers  comes  only  through  the  immense  gain  in  mathematical 
power  which  they  bring. 

71.  Extraction  of  square  root  of  polynomials.  This  process, 
from  which  a  method  of  extracting  the  square  root  of  numbers  is 
immediately  deduced,  may  be  performed  as  follows : 

EuLE.  Arrange  the  terms  of  the  polynomial  according  to  the 
powers  of  some  letter. 

Extract  the  square  root  of  the  first  term,  write  the  result  as 
the  first  term  of  the  root,  and  subtract  its  square  from  the  given 
polynomial. 

Divide  the  first  term  of  the  remainder  hy  twice  the  root 
already  found,  and  add  this  quotient  to  the  root  and  also  to  the 
trial  divisor,  thus  forming  the  complete  divisor. 

Multiply  the  complete  divisor  hy  the  last  term  of  the  root  and 
subtract  the  product  from  the  last  remainder. 

If  terms  of  the  given  polynomial  still  remain,  find  the  next 
term  of  the  root  hy  dividing  the  first  term  of  the  remainder  hy 
twice  the  first  term  of  the  root,  form  the  complete  divisor,  and 
proceed  as  before  until  the  desired  number  of  terms  of  the  root 
have  been  found. 


54  ALGEBRA  TO  QUADRATICS 

EXERCISES 

Extract  the  square  root  of  the  following : 

1.  a*  -  2  a^x  +  3  aH^  -  2  ax^  +  x\ 

Solution :  a*  -  2  a^x  +  3 a'^x'^  -  2 ax^  +  x*\a^  —  ax  +  x^ 

a* 

2a2-ax|  -  2  a^x  +  3  aH^  -  2  ax^  +  x* 
-  2  g^x  +     a^x^ 
2a^-2ax  +  x^\2 a^^  -  2 ox^  +  x* 
2  g^x^  -  2  ax8  +  x^ 

2.  1  +  x.  3.  1-x. 

4.  3x2  _  2x  +  x*  -  2x3  +  1.  5.  x*  -  6x3  +  13x2  -  12x  +  4. 

6.  x*  +  2/4  +  2x3?/  -  2x2/3  _  a;2?/2.  7.  9x*  -  12x3  +  34x2  -  20x  +  25. 

8.  49g4-42g36+37g262_i2a63+464.   9.  2g6- 2gc  -  2&c  H  g2  +  62  4.  c2. 

10.    W*102  +  v*U^  +  U>4v2  +  2  W3u2|0  _|_  2  t>8t«2w  +  2  W?3m2u. 

72.  Extraction  of  square  root  of  numbers.  We  have  the 
following 

Rule.  Separate  the  mcmher  into  periods  of  two  figures  each, 
heginning  o.t  the  decimal  point  Find  the  greatest  number 
whose  sqyiare  is  contained  in  the  left-hand  period.  This  is  the 
first  figure  of  the  required  root. 

Subtract  its  square  from  the  first  period,  and  to  the  remainder 
annex  the  next  period  of  the  number. 

Divide  this  remainder,  omitting  the  right-hand  digit,  by  twice 
the  root  already  found,  and  annex  the  quotient  to  both  root  and 
divisor,  thus  forming  the  complete  divisor. 

Multiply  the  complete  divisor  by  the  last  digit  of  the  root, 
subtract  the  result  from  the  dividend,  and  annex  to  the  remainder 
the  next  period  for  a  new  dividend. 

Double  the  whole  root  now  found  for  a  new  divisor  and  pro- 
ceed as  before  until  the  desired  number  of  digits  in  the  root 
have  been  found. 

In  applying  this  rule  it  often  happens  that  the  product  of  the  complete  divisor 
and  the  last  digit  of  the  root  is  larger  than  the  dividend.  In  such  a  case  we  must 
diminish  the  last  figure  of  the  root  by  unity  until  we  obtain  a  product  which  is 
not  greater  than  the  dividend. 


IRRATIONAL  NUMBERS  AND  RADICALS 


55 


At  any  point  in  the  process  of  extracting  the  square  root  of  a  number  before 
the  exact  square  root  is  found,  the  square  of  the  result  already  obtained  is  less 
than  the  original  number.  If  the  last  digit  of  the  result  be  replaced  by  the  next 
higher  one,  the  square  of  this  number  is  greater  than  the  original  number. 

There  are  always  two  values  of  the  square  root  of  any  number.  Thus  Vi  =  +  2 
or  —  2,  since  (+  2)2  =  (—  2)2=  4.  The  positive  root  of  any  positive  number  or 
expression  is  called  the  principal  root.  When  no  sign  is  written  before  the  radical, 
the  principal  root  is  assumed. 

EXERCISES 

Extract  the  square  root  of  the  following : 
1.  2.0000. 


Solution : 

2'.00'00'00'|1.414 
1 
2.4|1.00 
96 

281 1 400 
281 

2.824 1 11900 

-A 

11296 
604 

2.  96481. 

3.  56169. 

4.  3. 

5.  877969. 

6.  2949.5761. 

7.  5. 

8.  257049. 

9.  .00070128. 

10.  99. 

11.  69.8896. 

12.  .0009979281. 

13.  12. 

14.  49533444. 

15.  9820.611801. 

16.  160. 

73.  Approximation  of  irrational  numbers.  In  the  preceding 
process  of  extracting  the  square  root  of  2  we  never  can  obtain  a 
number  whose  square  is  exactly  2,  for  we  have  seen  that  such  a 
number  expressed  as  a  rational  (i.e.  as  .a  decimal)  fraction  does  not 
exist.  But  as  we  proceed  we  get  a  number  whose  square  differs 
less  and  less  from  2. 

Thus  1.2        =  1,  less  than  2  by  1. 

1.42      =  1.96,  less  than  2  by  .04. 
1.41^    =  1.9881,  less  than  2  by  .0119. 
1.4142  =  1.999396,  less  than  2  by  .000604. 


5Q  ALGEBRA  TO  QUADRATICS 

Though  we  cannot  say  that  1.414  is  the  square  root  of  2,  we  may 
say  that  1.414  is  the  square  root  of  2  correct  to  three  decimal 
places,  meaning  that 

(1.414)2  <  2  <  (1.415)2. 

74.  Sequences.  The  exact  value  of  the  square  root  of  most 
numbers,  as,  for  instance,  2,  3,  5,  cannot  be  found  exactly  in  deci- 
mal form  and  so  are  usually  expressed  symbolically.  By  means 
of  the  process  of  extracting  square  root,  however,  we  can  find  a 
number  whose  square  is  as  near  the  given  number  as  we  may  desire. 
We  may,  in  fact,  assert  that  the  succession  or  sequence  of  numbers 
obtained  by  the  process  of  extracting  the  square  root  of  a  number 
defines  the  square  root  of  that  number.  Thus  the  sequence  of 
numbers  (1,  1.4,  1.41,  1.414,  •  •  •)  defines  the  square  root  of  2. 

75.  Operations  on  irrational  numbers.  Just  as  we  defined  the 
laws  of  operation  on  the  fraction  and  negative  nimibers  (pp.  2-4), 
we  should  now  define  the  meaning  of  the  sum,  difference,  prod- 
uct, and  quotient  of  the  numbers  defined  by  the  seqilence  of  num- 
bers obtained  by  the  square-root  process.  To  define  and  explain 
completely  the  operations  on  irrational  numbers  is  beyond  the 
scope  of  this  chapter.  It  turns  out,  however,  that  the  number 
defined  by  a  sequence  is  the  limiting  value  of  the  rational  num- 
bers that  constitute  that  sequence,  that  is,  it  is  a  value  from  which 
every  number  in  the  sequence  beyond  a  certain  point  differs  by  as 
little  as  we  please.  We  may,  however,  make  the  following  state- 
ment- regarding  the  multiplication  of  irrational  nimibers :  In 
the  sequence  defining  the  square  root  of  2,  namely,  (1,  1.4,  1.41, 
1.414,  •  •  •)  we  saw  that  we  could  obtain  a  number  very  nearly 
equal  to  2  by  multiplying  1.414  by  itself.  In  general,  we  multi- 
ply numbers  defined  by  sequences  by  multiplying  the  elements  of 
these  sequences;  the  new  sequence^  consisting  of  the  products,  defines 
the  product  of  the  original  numbers. 

Thus     (1,  1.4,  1.41,  1.414,  •  •  )  (1,  1.4,  1.41,  1.414,  •  •  •) 
=  (1,  1.96,  1.9881,  1.999396,  •  •  •). 

The  numbers  in  this  sequence  approach  2  as  a  limit,  and  hence 
the  sequence  may  be  said  to  represent  2, 


IRRATIONAL  NUMBERS  AND  RADICALS  57 

76.  Notation.  We  denote  the  square  root  of  a  (where  a  repre- 
sents any  number  or  expression)  symbolically  by  Va,  and  assert 

or,  more  generally,  ^      _         , 

Va  •  V^  =  Va  •  b. 

Similarly,  Va  -^  V^  =  Va  -^  b. 

EXERCISES 

1.  Form  five  elements  of  a  sequence  defining  VS. 

2.  Form  five  elements  of  a  sequence  defining  V6. 

3.  Form  five  elements  of  a  sequence  defining  V6. 

4.  Form,  in  accordance  with  the  rule  just  given,  four  elements  of  the 
sequence  \/2  •  VS.    Compare  the  result  with  the  elements  obtained  in  Ex.  3. 

5.  Form  similarly  the  first  four  elements  of  product  V2  •  V6  with  the 
first  four  elements  obtained  by  extracting  the  square  root  of  10. 

77.  Other  irrational  numbers.  The  cube  root  and  higher  roots 
of  numbers  could  also  be  found  by  processes  analogous  to  the 
method  employed  in  finding  the  square  root,  but  as  they  are 
almost  never  used  practically,  they  will  not  be  included  here. 
It  should  be  kept  in  mind,  however,  that  by  these  processes 
sequences  of  numbers  may  be  derived  that  define  the  various 
roots  of  numbers  precisely  as  the  sequences  derived  in  the  pre- 
ceding paragraphs  define  the  square  root  of  numbers. 

The  Tith  root  of  any  expression  a  is  symbolized  by  Va.  Here 
n  is  sometimes  called  the  index  of  the  radical.  The  principle  for 
the  multiplication  and  division  of  radicals  with  any  integral 
index  is  given  by  the  following 

Assumption.  The  product  (or  quotient)  of  the  nth  root  of  two 
numbers  is  equal  to  the  nth  root  of  the  product  (or  quotient)  of 
the  members. 

Symbolically  expressed, 

Va  •  -y/b  =  VoT^, 

Va  H-  V^  =  Va  -hb. 


68  ALGEBRA  TO  QUADRATICS 

78.  Reduction  of  a  radical  to  its  simplest  form.  A  radical  is 
in  its  simplest  form  when  the  expression  under  the  sign  is  integral 
(§  11)  and  contains  no  factor  raised  to  a  power  which  equals 
the  index  of  the  radical ;  in  other  words,  when  no  factor  can  be 
removed  from  under  the  radical  sign  and  still  leave  an  integral 
expression.  We  may  reduce  a  quadratic  radical  to  its  simplest 
form  by  the  following 

Rule.  If  the  expression  under  the  radical  sign  is  fractional^ 
multiply  both  numerator  and  denominator  by  some  expression 
that  will  make  the  denominator  a  perfect  square. 

Factor  the  expression  under  the  radical  into  two  factors^  one 
of  which  is  the  greatest  square  factor  that  it  contains. 
■   Take  the  square  root  of  the  factor  that  is  a  perfect  square,  and 
express  the  multiplication  of  the  result  by  the  remaining  factor 
under  the  radical  sign. 

If  the  radical  is  of  the  nth  index,  the  denominator  must  be  made  a  perfect  nth 
power,  and  any  factor  that  is  to  be  taken  from  under  the  radical  sign  must  also  be  a 
perfect  nth  power. 

EXERCISES 


Reduce  to  simplest  form : 

1.  Vv. 

Solution : 

/T2          /12  .  5          /4  •  15 
\5=\    25    =\   25    ~- 

=  |Vl6. 
5 

2.  vi- 

3.  V32. 

4.  Vf. 

5.  V27. 

6.  VA. 

7.  V243. 

8.   V'250. 

9.  Vi  +  4. 

10.  Vi-|. 

11.  8V75. 

12.  tV2762. 
15.  V¥  +  |. 

13.  lV80a;8y*. 

14.  VtV  +  ^V 

16.  v-V^  +  ^V 

"■  xS- 

■        ^^  S 

■'■  xlf  ■ 

"M 

-->/!• 

«■f^/S■• 

-# 

-•  ^^m■ 

-  -Vi^- 

IRRATIONAL  NUMBERS  AND  RADICALS  69 


26.   Vx^  -  2  x2y  +  X2/2.  27.  V6x8-20x2  +  20x. 


28       M^-^«'^  +  «  29       /2a;«-I2x2:+l8x 

„Q       /2 gs  -  8 a2  +  8 g  „-        ja^  +  a% -~ab' 

'    \8x-8x2  +  2x3'  *    \  9(g-&.) 


g62  _  63 


79.  Addition  and  subtraction  of  radicals.  Radicals  that  are  of 
the  same  index  and  have  the  same  expression  under  the  radical 
sign  are  similar.  Only' similar  radicals  can  be  united  into  one 
term  by  addition  and  subtraction.  We  add  radical  expressions  by 
the  following 

Rule.    Reduce  the  radicals  to  he  added  to  their  simplest  form. , 
Add  the  coefficients  of  similar  radicals  and  prefix  this  sum 
as  the  coefficient  of  the  corresponding  radical  in  the  result. 

A  rule  precisely  similar  is  followed  in  subtracting  radical 
expressions. 

EXERCISES 

Add  the  following : 

1.    V27,  \/48,  and  V75. 
Solution : 


V27  =  - 

n/  9  •  3  =    3  V3 

V48=- 

v/I6.3=    4V3 

\/75=^ 

v/25.3=    6V3 

Sum  =  12  \/3 

2.  V3  +  2V3.  3.  8V7-3V7.  4.  ay/x-hy/x. 

5.  g  +  2Vg  +  3\/a  +  2  Vl6g  -  \/27g. 

6.  3  V8  +  4  V^  ~  5  V50  -  7  V72  +  6  V98. 

7.  8Vg+5\^-7Vg  +  4Vg-6Vx-3 Vg. 

8.  7  V4x  +  4  Vox  +  3  ViSx  -  5  V36x  -  2  V80x. 

9.  Vg^^  +  Vl6g  -  106  +  Vgx2  -  6x2  _  V9(g  -  6). 


10.  4  Vo^  -  3  y/¥x  +  2  Vc^  +  Vd2x  -  2  V(6  +  dfx. 

11.  6  Vx  +  3  V2x  -  5  V3x  -  2  V4x  +  Vl2x  -  \/l8x. 


60  ALGEBRA  TO  QUADRATICS 

80.  Multiplication  and  division  of  radicals.   For  these  pro- 
cesses we  have  the  following 

EuLE.    Follow  the  usual  laivs  of  operation  (§  10),  using  also 
the  assumption  of  §  77. 

Beduce  each  term  of  the  result  to  its  simplest  form. 

"  The  operations  of  this  section  are  limited  to  the  case  where 
the  radicals  are  of  the  same  index.  Radicals  of  different  indices 
as  Vs  and  V^  must  first  be  reduced  to  the  same  index.    See  §  87. 

EXERCISES 

1.  Multiply  V2  -  V3  by  V2  -  VS. 
Solution :  V2  -  V3 

V2-V8 


2  _  V6  -  Vl6  +  V2i 
=      2-V6-4  +  2V6 

=  -  2  +  V6. 


2.  Divide  ^_tZ_  by  v^  +  \/y. 
Vxy 
x-\-y 

y/xy           (  Vx2  -  Vxy  +  Vy2)  (  Vx  +  y/y)  * 
Solution :  — — ——  = — ^ ^ 

Vx  +  y/y  y/xy  (  Vx  +  Vy) 

_  Vx2       Vxy       Vy2 
Vx2/       Vxy       V^ 


<^--</!- 


Carry  out  the  indicated  operations  and  simplify : 
3.  VlO .  V5.  4.  V^  .  ■^. 

5.  V28 .  V7.  •  6.  Vf  ^  Vf- 

7.  (a -6x^1  8.  VS-Vff. 

9.  (  V7  -  V3)  (  V3  -  V2).  10.  (-  1  +  V3)^ 

11.  (6  V3  +  V6)  (6  V2  -  2).  12.  y/lTc  •  VTOc. 

*  Since  as  +  63=  (a  +  ft)(a«  -  a6  +  ft*),  if  o=  V^,  6=  Vy  we  have 


IRRATIONAL  NUMBERS  AND  RADICALS  61 

13.  (a  +  6-V^)(Va+V&).  14.  (8  +  3  Vs)  (2  -  VB). 


15.  VVx  +  Vy  •  V Vx  -  s/y,  16.  V6  +  2  V6 .  V 6  -  2  V5. 


17.  Vx  +  Vjc2  -  1 .  Vx  -  Vx2  -  L  18.  (x2  +  y2)  ^  (a;  ^  _,.  y  ^), 


19.  --V^.  20.  (4Va-V3^)(V^  +  2V3x). 


X 

21.  VS.Jp. 
\4a 


25.  (^.^Ij.  36.V^.Vg 


27,  ^*-^ 


+  66 
y  '    \x/  \ ox^  —  to2 

:fl^.(.t^-,%).  28.  (5-l^).(l    -    1). 

Vxy  \y     ^1     VVx     Vy/ 

29.  (2  V6  -  Vl2  -  V2i  +  Vis)  V2. 

30.  (3\/8  +  Vl8+V60-2\/72)V2. 

31.  (5  v^  -  4  V32  +  3  V60  -  3  V54)  V3. 

32.  ( V9^T6  +  3  A^)(V9x  +  6  -  3  \^). 

33.  [(V7  4  V3  +  Vl0)(V7  +  V3-Vi0)7. 

34.  (2  V30  ..  3  V6  +  5  V3)  (  V8  +  V3  -  Vs). 

35.  (2V^  +  V8-Vl2)(^V30-fV3+V2). 


37.  Find  the  value  of  ^  V2i  -  V|  +  2  Vs  -  V6- V3  4- V5  to  three 
decimal  places. 

81.  Rationalization.  The  process  of  rendering  the  irrational 
numerator  (or  denominator)  of  a  fractional  expression  rational 
without  altering  the  value  of  the  fraction  is  called  the  rationaliza- 
tion of  the  numerator  (or  denominator)  of  the  fraction. 

This  is  usually  accomplished  by  multiplying  both  numerator 
and  denominator  of  the  given  fraction  by  a  properly  chosen 
radical  expression  called  the  rationalizing  factor. 


62  ALGEBRA  TO  QUADRATICS 

The  principles  in  accordance  with  which  this  rationalizing 
factor  is  selected  are  the  following: 

Principle  I.  Since  {a  ■i-b)(a  —  b)=a^  —  b%  the  rationalizing 
factor  of  -sfx  ±  Vy  is  sjx  hF  Vy. 

Principle  II.  Since  {a^  —  a'b-\-lF)  (a  +  5)  =  a'  +  h^,  the  ration- 
alizing factor  of  -sfx  -\-  Vy  is  V^  —  V^  +  Vy^  and  conversely. 

Since  (a^  -\-  ah  -\-  If) (a  —  b)=a^  —  b^,  the  rationalizing  factor 
of  -y/x  —  -Vy  is  V^  +  Vxy  +  Vy^,  and  conversely. 

EXERCISES 

^Nationalize  the  denominators  of  the  following : 
-    Va  +  Vx 
Va  —  Vx 
Solution :  By  Principle  I  the  factor  which  will  render  the  denominator 

rational  is  Va  +  Vx. 

Va  +  Vx      Va  +  Vx   Va  +  Vx     a  +  x  +  2  Vox 


Thus 
2. 


Va—Vx      Va—Vx    Va  +  Vx 
1 


2+V2  +  V3 
Soltttioii :  This  problem  requires  a  twofold  rationalization. 

1  (2+V2)-V3 

2  +  V2  +  V3~  [(2+ V2)  +  V3][(2 +\^)  -  V3] 
^2+-v^-V3^      2+V^-V3 
~  (2  +  V2y-S  ~4  +  4V2  +  2-3 
_2+V2-V3_(2  +  v^-V3)(3-4\^ 

3  +  4v^     ~      (3  +  4V2)(3-4v^) 
_6  +  3-v^-3V3-8\/2-8  +  4V6 
""  9  -  16  .  2 

2  +  6v^+3V3-4V6 


^2-V3 
Solution : 


^2-2  +  ^6  +  ^^     Vi  +  v'e  +  v^ 


^-.■^S      ^-^3    ^2+^6+V3-«  2-3 

=  ^(^  +  V6  +  ^9). 


IRRATIONAL  NUMBERS  AND  RADICALS  63 

5    7-V5  g    V3+V2 


2+V3  3+V6  V3-V2 


8  /-  3, 


V3+V2  V2-V4  V2+3Vi 

10.  ^/^«±^.  11.     I^±^^. 

\a-y/x  \a-Va2-l 

12  2V6  ^2    1+3V2-2V3 

*   V2+V3+V6  '    V6_}.V3+V2 

--  2\/T6  -g    Va  +  ic  +  Va  —  x 


V3  +  V5  +  2  \/2  Va  +  x  -  Va  -  x 

16  2  j^    V6-V5-V3  +  V2 

Va  +  1  +  Va-  1  V6  +  V5  -  V3  -  V2 

18.  Show  that  ^^  ~  "^  =  -  .10  •  ■  -. 

V2-fV3 

19.  Show  that  ^^^-^^  =  17.48 .... 


V8- 

-V7 

20 

V(l  +  a)(H-6)- 

-V(i 

-a)(l 

-&) 

21. 

V(l  +  a)(l  +  6)  +  V(l 
Show  that  ^^+^^ - 

-a)(l 

-h) 

V6_ 

=  .168. 


V3  +  V5  +  V6  -  V5 

82.  Solution  of  equations  involving  radicals.  We  prove  the 
important 

Theorem.  WTien  an  equation  in  x  is  multiplied  hy  an  expres- 
sion in  Xy  the  resulting  equation  has,  in  general,  solutions  which 
the  first  one  did  not  possess. 

Let  ^  =  0 

represent  an  equation  contaming  x  which  is  satisfied  by  the 
values  X  =  a,  by  ■  -  n.  Let  5  be  an  expression  which  vanishes 
when  X  =  ay  p,  •  '  V.    Then  the  expression 

is  satisfied  not  only  when  x  =  a,  bj  •  -  j  n,  but  also  when 
x  =  a,l3,-';v.      . 


64  ALGEBRA  TO  QUADRATICS 

Example.    The  equation  x  —  2  =  0 

has  X  =  2  for  its  only  solution,  while  the  equation  , 

(X  -  2)  (X  -  3)  =  0 
has  in  addition  the  solution  x  =  3. 

If  in  the  course  of  a  problem  it  is  necessary  to  multiply  an 
equation  by  any  expression  involving  the  variable,  the  solutions 
of  the  resulting  equation  must  be  substituted  in  the  first  one  to 
ascertain  if  any  solutions  have  been  introduced  which  did  not 
satisfy  the  original  equation.  Solutions  which  have  been  intro- 
duced in  the  process  of  solving  an  equation,  but  which  do  not 
satisfy  the  original  equation,  are  called  extraneous  solutions. 

It  may  be  shown  in  a  similar  way  that  raising  the  equation  in  Xj 

A  =B, 
to  any  power  introduces  extraneous  solutions. 

EXERCISES 


1.  Solve  Vaj  +  19  +  Vx  +  10  =  9. 

Solution :       '^  Vx  +  19  =  9  -  Vx  +  10.  

X  +  19  =  81  +  X +  10-  18  V»Tl0. 
-72=-18  Vx  +  10. 
4  =  Vx  +  10. 
16  =  x  +  10. 
x  =  6." 
Check :         V6  +  19  +  V6  +  10  =5  +  4  =  9. 

2.  Solve  Va;  +  19  -  Vx  +  10  =  -  9. 

Vx  +  19  =  -  (9  -  VxTlO). 
Simplifying,  we  get  x  =  6. 


Check :         V6  +  19  -  V6  +  10  =+6-4  =  1  9^-9. 

Thus  our  result  satisfies  only  the  equation  which  was  introduced  in  the 
course  of  solving  the  problem,  and  is  extraneous.  The  Original  equation 
has  no  solution. 

Solve  and  check,  noting  all  extraneous  solutions : 

3.  VSx  -1=5.  •  4.  Vi«  -8  =  2. 

5.  V2x  +  V3x  =  1.  6.  Vx  +  V3x  =  2. 

7.  V5x-7=v'4x  +  3.  8.  5Vx-7  =  3v^-l. 


IRRATIONAL  NUMBERS  AND  RADICALS 


65 


9.  Vx  +  1  +  Vx  +  2  =  3. 
11.  2\^-V2x  =  2+v^. 
13.  V37  -  7  V6x^=^  =  4 


10.  Vl3  4-4v^^^ 


6. 


12.  Vx  +  4  +  \^+T=L 


14.  Vx2  -  7  x  +  19  =  v^^^. 


15.  7  +  Va;2-lla;  +  4  =  jc. 


17.  a;  —  Vax  (1  +  x)  +  1  -  x  =  1. 
19.  |(7V5  +  6)-5=|(3v^-l). 
I  4 


16.  8  +  V(x  -  10)  (X  -  6)  =  X. 
18.  V2  (X  +  1)  +  V2x  +  15  =  13. 
20.  2\/3  +  3V2x  =  3V2  +  2V3a 


21. 


6  4- Vx      8 


23.  V7  X  +  2  = 


25.  V2X-1 


-Vx 

6x  +  6 
V7X4-2' 

2  (x  -  3) 
V2x-10' 


22. 


24. 


4. 


5  + v^ 
5-Vx 
a  —  Vto      26 


3  Vox 


a  +  V6x      2  6  +  3  Vox 
6x+  10 


26.   V9  X  +  10  = 


27. 


VTT- 


VT 


1  +  VI -X     1  -  VIT^ 

29.  X  —  ax :  Vx  =  Vx :  X. 

30.  2v^^T2  +  V^T2  =  i^^±^. 

V8x  +  8 

1  +  2V3X-6       11  +  2  V3x- 5 


V4X+9 


28.  Va  -  X  +  V6  -  X  = 


V6 


31. 


1  +  3V3X-6      11  +  5V3X-5 


32.  V9x  +  7  +  V4X  +  1  =  V25x  +  14. 

33.  Vx  +  15  +  Vx  -  24  -  Vx  -  13  =  Vx. 


34.  Vx  -  7  +  Vx  -  2  -  Vx  -  10  =  VxT5. 

35.  (VS-7)(Vx-3)  =  (Vx-6)(Vx-5). 

36.  (a  +  \^)  Vx:(6-Vx)  Vx  =  a  +  1:6-L 

37.  (4V5-7):(5V^-6)  =  (V^-7):(V^-6). 

38.  (  Va  V6  —  V6  Va) Vx  =  a V6  Vx  —  6 VaVx. 


CHAPTEE  VII 

THEORY  OF  INDICES 

83.  Negative  exponents.   We  have  already  seen  (§  16)  that 

a^-a"*  =  «"  +  "'  (1) 

when  n  and  m  are  positive  integers.  We  now  assume  that  this  law 
still  holds  when  one  or  both  of  the  numbers  m  and  n  are  negative 
or  fractional. 

If  we  let  a-m  =      , 

a'" 


then 


since  the  law  (1)  holds  when  n  and  m  are  any  integers.    This 
notation  may  be  expressed  verbally  as  follows : 

Principle.  A  factor  of  numerator  or  denominator  of  a  frac- 
tion may  be  changed  from  the  numerator  to  the  denominator, 
or  vice  versa,  if  the  sign  of  its  exponent  he  changed. 

84.  Fractional  exponents.  Since  (p.  57)  Va  •  Va  =  a,  it  is 
natural  to  devise  a  notation  for  Va  suggested  by  the  law  (1). 

If  we  let  Va  =  a*, 

we  have  Va  •  Va  =  a*  •  a^  =  a'  "^  *  =  a^  =  a. 

Furthermore,  if  we  let   -y^  —  an 
it  would  be  consistent  with  law  (1)  to  write 

1  11  11  s 

(a")'' =  a«  .  a»  =  a"    "  =  «"'. 
This  notation  we  shall  assume  in  general.    Thus 

66 


THEORY  OF  INDICES  67 

With  the  adoption  of  this  notation  we  can  attach  a  meaning 
:o  any  real  number  with  any  rational  number  for  its  exponent. 
This  notation  may  be  expressed  verbally  in  the  following 

Principle.  The  numerator  of  a  fractional  exponent  indicates 
a  power,  the  denominator  a  root. 

85.  Further  assumptions.  The  operation  of  multiplication  is 
subject  to  the  following  laws  of  exponents : 

I.   Commutative  law  of  rational  exponents : 

(^a'^y  =  a""-  =  a'--''  =(a'-y. 

II.  Associative  law  of  rational  exponents : 

(yj  =  a'?'- -^  =  ««"•«  =  (a'^y. 

The  laws  of  operation  (§  10)  defined  for  integral  values  of  the 
symbols  we  also  assume  when  the  symbols  are  expressions  with 
rational  exponents. 

86.  Theorem.  a''lf=  {ctby,  where  r  is  any  rational  number  as  _. 

We  raise  both  sides  of  the  equation  a'b''  =  (aby  to  the  g-th 
power  separately  and  show  that  the  results  are  equal. 

Since  r  =  -> 

l(abyY  =  [(abyj  =  {aby  =  aPbP. 

p  p  p  p      p  p  p  p 

Also  (or by  =  (a«55)   =  (a^^»«)(a«6«)  •  •  •  {a'^^) 


q  terms 

p     p 

P        P       P            P 

q  terms 

g  terms 

=iJm' 

==  a^hP, 

{ariry 

=  l(abyj. 

ing  the  qih 

root  and  taking  the  principal 

root, 

we 

obtain 

a'b"- 

=  (aby. 

• 

68  ALGEBRA  TO  QUADRATICS 

EXERCISES 
yl.  Express  in  simplest  form  with  positive  exponents : 

,  ,  36a-26-ic-5 

(a) 

9a26-2c-i 

Solution :  By  Principle,  §  83,  ^^^"^^~^^"^ 
9a26-2c-i 

_46-i&+2c-ic+3 
~  a2-a2 

^46c 
~  a*  ' 

(b) (c)  — - 


^  ^    (xi2/l)-i  '  35x-2?/6z-4* 

3a-i6-2    6a2x-i  gft^c  Va6v^ 

4X-22/-4'  56-ic2'  ^'^  v'^&-^aJ62c*' 

,..     5    11     / — I — \ — ;  /.,  V   4x-«y-3     15a'263-m 

^"•^  •  '  5  a- 4 6-"'      14x«2/«-3 

•  2.  Arrange  in  order  of  magnitude  the  following : 

(a)  Vi,   71,   7|. 

Solution :  We  first  ask.  Is  (f  )i  >  (|)^  ? 

Raise  both  numbers  to  the  sixth  power. 

We  obtain  (|)3    and    (f)2, 

or  If    and    f, 

er  2if    and    2|. 

Thus  Vl>7l- 

Now  compare  (f)^    and    (|)io 

Raise  both  numbers  to  the  fourth  power. 
We  obtain  .  (|)2    and    |, 

or  1|    and     If. 

Thus  VI  >  Vl. 


THEORY  OF  INDICES  69 

Now  compare  (§)'    and    (|)^. 

Raise  both  numbers  to  the  twelfth  power. 
We  obtain  (f)*    and     (|)3, 

or  fl    and    \\^-, 

or  6yV    and    6f|. 

Thus  -s/i>\^h 

The  order  of  magnitude  is  then  Vli  VI?  Vf  • 

(b)  V^,  VI,   </h  (c)  4^,  V8,  -Ws. 

(d)  Vli,   VI-  (e)  V3,   ^5,   Vl5. 

3.  Perform  the  indicated  operations, 
(a)  \/2  .  V3. 

Solution :  ^2  •  V3  =  2'  •  3^  =  2^  3^ 

=  (22 .  38)i  =  V'4^  =  v^lOS. 

V3 

V5  .  ''V3.V2 

87.  Operations  with  radical  polynomials.  These  operations 
follow  the  rules  for  the  same  operations  previously  given,  pro- 
vided the  assumptions  and  principles  of  §§  83-86  are  observed. 

EXERCISES 

1.  Divide  x^  —  y^  by  ^/x  —  y/y. 

2.  Extract  the  square  root  of  4  x  -  12  x^  2/3  +  9  2/5  +  32  x^  -  48  yi  +  64. 
3x  +  3x-i-6 


3.  Simplify 


xi-  3x2  +  3x-^-x-i 


4.  Divide  J-Va^  -J-V^  by  sJ-> 

a'2      52 

5.  Extract  the  square  root  of 1 2. 

6.  Multiply  -  3x-5  +  2^  by  ^  -  ?^. 

X*  x^         6-1 

7.  Extract  the  square  root  of 

-ix2y-2-  i^yx-i  +  ^y2x-2-^xy-^  +  25?. 
49  2  Id  7  7 

8.  Multiply  V^  -  x»  +  x^ f-  Vx^  -x+v^-lby  Vx  +  1. 


QUADRATICS  AND  BEYOND 

CHAPTEE  VIII 
QUADRATIC  EQUATIONS 

88.  Definition.  An  equation  that  contains  the  second  bnt  no 
higher  power  of  the  variable  is  called  "a  quadratic  equation.  The 
most  general  form  of  the  quadratic  equation  in  one  variable  is 

ax'^  +  hx-{-c  =  0,  (1). 

where  we  shall  always  assume  a,  b,  and  c  to  represent  rational 
numbers,  and  where  a  =^  0.  Every  quadratic  equation  in  x  can 
be  brought  to  this  form  by  transposing  and  simplifying. 

89.  Solution  of  quadratic  equations.  The  solution  of  a  quad- 
ratic equation  consists  in  finding  its  roots,  that  is,  the  numbers 
(or  expressions  involving  the  coefficients  in  case  the  coefficients 
are  literal)  which  satisfy  the  equation. 

The  common  method  of  solving  a  quadratic  equation  consists  in 
bringing  the  member  of  the  equation  that  involves  the  variable 
into  the  form  of  a  perfect  square,  i.e.  into  the  form 

x^-\-2Ax  +  A\ 

For  example,  let  us  solve 

Transpose  8,  x^-^2x  =  S. 

If  now  we  add  1  to  both  sides  of  the  equation,  the  left-hand 
member  will  be  a  perfect  square, 

x^-\-2x-{-l  =  9. 
70 


QUADRATIC  EQUATIONS  71 

Express  as  a  square,       (x  +  1)^  =  9. 

Extract  the  square  root,     x  -\-l  =±S. 

Transpose,  ic  =  —  4  or  2. 

Both  —  4  and  2  satisfy  the  equation,  as  we  see  on  substituting 

them  for  x.    Thus 

(-4)^  +  2(-4)-8  =  0, 

and  2^  +  2  •  2  -  8  =  0. 

Consider  now  the  general  case. 

Let  us  solve  ax'^  -{-  bx  -\-  c  =  0. 

Transpose  c,  ax^  -\-  bx  =—  c. 

b              c 
Divide  by  a,  x^  -^  -  x  = 

/  bV 
Add    (  77—  I  to  both  members  to  make  the  left-hand  member  a 

\2aJ 

perfect  square, 

b  b''   _      c        b""    _-4.ac-\-b^ 

,a         4a^  a       4a^  4  a^ 

4  ac 


/      by   b^-4 

Express  as  a  square,  (  x  +  — —  I  =  — —^ 
Extract  the  square  root. 


b  V^^  —  4  ac 


x-\-w-  =  ± 


2a  2a 

2a 


Transpose,  x  = ^^ *  (1) 

The  roots  are 


-b-\-  V^^  -  4  ac         _  -b  -  -s/b^  -4.ac 
Xi  —  „  f  X2  —  „ 

2  a  2a 

That  the  equation  can  have  no  other  roots  appears  from  §  96. 

*  This  expression  for  the  roots,       _         /         — 

X  = » 

2a 

may  be  used  as  a  formula  for  the  solution  of  a  quadratic  equation. 

Thus  to  solve  the  equation  2a;2-3a;  —  6=0 

we  may  substitute  in  the  formula  a  =  2,  6  =  -  3,  c  =  —  6,  and  obtain 

3±\^+48     3±V57 


Thus  Xi-. 


4  4 

3+V57  3-V57 


72  QUADRATICS  AND  BEYOND 

One  should  verify  the  fact  that  both a-iid 

—  satisfy  (1)  and  are  consequently  roots  of  the 


2a 

equation.  They  are,  in  geueral,  distinct  from  each  other.  For 
particular  values  of  the  coefficients  to  be  noted  later  (§  98)  the 
roots  may  be  equal  or  complex  (i.e.  of  form  a  -\-  /3  V— 1,  where 
a  and  yS  are  ordinary  rational  or  irrational  nimibers). 

We  may  sum  up  the  process  of  solving  a  quadratic  equation  in 
the  following 

EuLE.    Write  the  expression  in  the  form  aa^  -\-  hx  -\-  c  =  0. 

Transpose  the  term  not  involving  x  to  the  right-hand  side  of 
the  equation. 

Divide  both  sides  of  ^he  equation  by  the  coefficient  of  a^. 

Add  to  both  members  the  square  of  one  half  of  the  coefficient 
of  X,  thus  making  the  left-hand  member  a  perfect  square. 

Rewrite  the  equation,  expressing  the  left-hand  member  as  the 
square  of  a  binomial  and  the  right-hand  member  in  its  simplest 
form. 

Extract  the  square  root  of  both  members  of  the  equation,  not 
omitting  the  ±  sign  in  the  right-hand  member. 

Transpose  the  constant  term,  leaving  x  alone  on  the  left-hand 
side  of  the  equation.  The  two  values  obtained  on  the  right-hand 
side  by  taking  the  +  and  —  signs  separately  are  the  roots  sought. 

Check  by  substituting  the  solutions  in  the  original  equation, 
which  should  then  reduce  to  an  identity. 

90.  Pure  quadratics.  A  quadratic  equation  in  which  the  coeffi- 
cient of  the  term  in  x  is  zero  is  often  called  a  pure  quadratic.  Its 
solution  is  found  precisely  as  in  the  general  case,  excepting  that 
we  do  not  need  to  complete  the  square.    Thus  let  us  solve 

ax^  -f-  c  =  0. 

Transpose  c,  ax^  =—  c. 

Divide  by  a,  ic*  =  —  -. 


QUADRATIC  EQUATIONS  73 


Extract  the  square  root,            ^  =  ±  _,' 
The  roots  are  Xi  =  +  -v/—  -?  x^  =  ~  a/ 


EXERCISES 

Solve  and  check  the  following : 
1.  3x2 -6x- 10  =  0. 

Solution :   Transpose  10,  3  x^  —  6  x  =  10. 

.  Divide  by  3,  x2  -  2  x  =  J/- 

Add  the  square  of  \  the  coefficient  of  x,  i.e.  1,  to  both  sides, 

x2_2x  +  l  =  -Lo  +  l  =  Y- 
Express  as  a  square,  (x  —  1)2  =  -LS-. 

Extract  the  square  root,  x  —  1  =  ±  "V^- 

x  =  i±V¥. 

Check:   3(l±^'-6(l±^)-10  =  0. 


3±6^|  +  |-6:F6Vir_io  =  0. 


/13 


2.  8x2  +  2x  -3  =  0. 

Solution :   Transpose  3,  8  x2  +  2  x  =  3. 

Divide  by  8,  x2  +  i  x  =  f . 


Add  the  square  of  \  the  coefficient  of  x,  i.e.  ^^j,  to  both  sides, 

^'  +  \^  +  -h  =  \  +  iz  =  il' 

Express  as  a  square. 

(^  +  \Y  =  f  f- 

Extract  the  square  root 

X  +  1  =  ±  |. 

^=±'J-\ 

=  -lov\. 

Check:   8. (1)2  +  2. 1- 

3  =  1  +  1-3  =  0. 

8(-f)^  +  2(-f)- 

-3  =  11-1-3  =  1-1-3=1-1 

3.  x2-ax  =  0. 

4.  x2  =  169. 

5.  x2-ix  =  i 

6.  fx2  =  560. 

7.  x2  +  x-l  =  0. 

8.  19x2  =  5491. 

9.  3x2-7x  =  16. 

10.  x2  =  .074529. 

11.  3x2+  ll  =  5x. 

12.  x2  -  tl  X  =  1. 

13.  x2  +  X  -  66  =  0. 

14.  20x2  + x  =  12. 

1  =  0. 


74  QUADRATICS  AND  BEYOND 

15.  7a;2  +  9x  =  100.  16.  6x2  +  6x  =  66. 

17.  14x2  -  33  =  71x.  18.  5x2  +  13  =  14x. 

19.  x2  -  8x  +  15  =  0.  20.  91  x2  -  2 X  =  45. 

21.  x2  +  2 X  -  63  =  0.  22.  x2  -  6x  +  16  =  0. 

23.  x2  -  lOx  +  32  =  0.  24.  6x2  +  26^  =  25  ix. 

25.  6x2  -  13x  +  6  =  0.  26.  15x2  +  527  =  178x. 

27.  2x2  +  15.9  =  13.6 X.  28.  (x  -  1)2  =  a(x2-  1). 

29.  a'^{b-x)^=h^{a-x)^.  30.  13x2-19  =  7x2  +  5. 

31.  ax2  -  (a2  +  l)x  +  a  =  0.  32.  (a  -  x)(x  -  6)  =  -  ab. 

33.  14  x2  +  45. 5  X  =  -  36. 26.  34.  a^  {a  -  x)2  =  b'^{b-  x)2. 

35.  (a-x)2+(x-6)2  =  a2  +  62.  35.  (a -x)(x-6)  =  (a -x)(c-x> 

37.^'  =  ^.  38.  2x  +  l  =  3. 
b       d  X 

33    15x^810^  40.  x2  +  ?  =  50. 

2  3x  7 

-  -    2  X      1050  yio«  +  a^,^  +  a^      5 

41.  —  = 4<c. 1 = -. 

3  7x  6  +  xa  +  x2 

43    ^  +  ^^  :^  2x  +  l  44    x3- 10x2  +  1  ^  ^  _  g 
■x  +  3       x+6*  'x2-6x  +  9 

.^    ax  +  b      mx  —  n  ac    ^^         1^  4 

45.  = 46. =  0. 

6x  +  anx  —  m  xx  —  2x-3 

47.  -— ^—  =  -.  48.  — - —  + =  X  -  1. 


49. 


te2  —  mx  +  n      n  9  2  x  —  3 

x-2    _  3  (8  -  x)  gQ    (a-x)3  +  (x-6)8_  a^-b^ 

}x+  14~    28-x  '  *  " 


51.'^— +  ^-^  =  12.  52. 

2x-  3      X-  I 

53    5  +  X     '8  -  3  X  _    2  X  -^ 

'  3  —  X  X  x-2 

55.   (^'=8(^-15.  56.5^ 

\x  -b)         \x-bj  9 

57.  a2  -  x2  =  (a  -  X)  (6  +  c  -  x). 


(a- 

-X)- 

-(X- 

-6) 

a 

+  &■ 

16- 

■  X 

2(x 

-11) 

x-4 

4 

X  • 

-6 

12 

2x- 
X  — 

T* 

3x- 

X  - 

4-1 
-3 

5x 

X 

-14 
-4 

5x- 

^V 

3x- 

-1_ 

?4 

■X  — 

58.  i=  +  ^ =  ^. 

59    2_^±^  +  J?-  =  ^Ili  +  ^Z^. 
18         x  +  4  4  6 

60.  {X  -  a  +  6)  (X  -  a  +  c)  =  (a  -  6)2  -  x^. 


QUADKATIC  EQUATIONS  75 

91.  Solution  of  quadratic  equations  by  factoring.  When  the 
left-hand  member  of  an  equation  can  be  factored  readily,  this  is 
the  most  convenient  method  of  solution.  It  also  illustrates  very 
clearly  the  meaning  and  property  of  the  roots  of  the  equation. 

Example.    Solve  x^  -}-  2  x  —  15  =  0. 

Factor  the  left-hand  member,     (x  +  5)  (x  —  3)  =  0. 

The  object  in  solving  an  equation  is  to  find  numbers  that  substituted  for 
the  variable  satisfy  the  equation.  But  since  zero  multiplied  by  any  number 
is  zero  (p.  3),  any  value  of  x  which  causes  one  factor  of  an  expression  to 
vanish  makes  the  whole  expression  vanish.  If  in  this  case  x  =  3,  our  equa- 
tion in  factored  form  becomes 

(3  +  5)  (3  -  3)  =  (3  +  5)  •  0  =  0 
and  is  satisfied.    If  we  let  x  =  -  5,  the  other  factor  becomes  zero,  and  the 
equation  reduces  to  the  identity 

(5_5)(6-3)  =  0(5-3)  =  0. 

Thus  the  numbers  3  and  —  6  are  solutions  of  the  equation. 

92.  Solution  of  an  equation  by  factoring.  We  have  imme- 
diately the 

EuLE.  Transpose  all  the  terms  to  the  left-hand  member  of  the 
equation. 

Factor  that  member  into  linear  factors. 

The  values  of  the  variable  that  make  the  factors  vanish  are 
roots  of  the  equation. 

EXERCISES 

Solve  and  check  the  following : 

1.  6x2  +  x  =  15. 

3.  6x2 -X- 6  =  0. 

5.  13x2-38x  =  3. 

7.  x2-40x  + 111  =  0. 

9.  x2-18x- 208  =  0. 
11.  x2-3ax-4a2  =  0. 
13.  (x2  _  1)  +  (X  -  1)2  =  0. 

15.  (2x  -  l)(x  -I-  2)  -f  (X  -  1) (X  -  2).=  -  4. 

16.  (7x  -  l)(x  -f  3)  -  (4x  -  3)(x  -  1)  =24. 


2. 

6x2-f-7x  =  3. 

4. 

5x2-17x+6  =  0. 

6. 

2x2 -6x- 26  =  0. 

8. 

13x2 -40x  + 3  =  0. 

10. 

3x2 -26x  + 36  =  0. 

12. 

(X  -  a)2  -  (x  -  6)2  =  0. 

14. 

(3x_6)2_(9x-M)2  =  0. 

76  QUADRATICS  AND  BEYOND 

17.  (3x  +  1)  (X  +  1)  -  (4a;  +  3) (x  -  1)  =  -  2. 

18.  (X  -  a)(4ax  -  6)  +  (x  -  6)(4ax  -  6)  =  0. 

19.  3x-4Vx^^  =  2(x  +  2). 


Solution :  Transpose,  x  —  4  =  4  Vx  —  7. 

Square,  x2  -  8  x  +  16  =  16  x  -  112. 

Transpose,  x2  -  24  x  -f  128  =  0. 

Factor,  (x  -  16)  (x  -  8)  =  0. 

The  roots  are  x  =  16.    x  =  8. 


Check :   3-16-4  Vl6  -  7  -  2  (16  +  2)  =  48  -  12  -  32  -  4  =  0. 
3-8-4  VS~^  -  2(8  +  2)  =  24  -  4  -  20  =  0. 

In  the  following  examples,  as  always,  the  quadratic  equation  should  be  solved 
by  factoring  when  possible.  Recourse  to  the  longer  but  sure  method  of  completing 
the  square  is  always  available. 

When  an  equation  is  cleared  of  fractions  or  squared  in  the  process  of  bringing 
it  into  quadratic  form  (1),  §88,  extraneous  solutions  may  be  introduced.  The 
results  should  be  verified  in  every  case  and  extraneous  solutions  rejected. 

20.  a  +  Va2  -  x^  =  x.  21.  VxT~5  =  x  -  1. 


22.  2x-^V2x-l  =  x  +  2.  23.   1-  6x  +  V5(x  +  4)  =  0. 

24.  Vll-x  +  \/x-2  =  3.  25.  V2X  +  1-2  V2x  +  3  =  1. 


Hint.  Square  twice. 


26.  Va(x-6)  +  V6(x-a)  =  x. 


27.  VxTS  +  V2X-3  =  6.  28.  2  V3  +  X  -  4  VS^^  =  V60. 


29.  Vl  +  ox  -  Vl  -  ax  =  X.  30.  V5x- 1  -  V8-2x  =  Vx^. 


31.  Vx  +  7-V5x-2  =  3.  32.  x  -  10  =  |(x  -  1)-V2x-1. 


33.  Va2"^^  +  V&2  +  X  =  a  +  &.         34.  V4  -  x  +  V6  -  x  =  V9  -  2; 


X 

X         \b 


35.  (a2-62)(x2+l)  =  2(a2  +  62)x.  36.  x  +  2  a V2(a2  +  62) - x  =  3a2  +  62. 

37.  xV^:r2  +  2V^T2zzV^H^.  38.  -J^-^  +  \f-^  =  2- 

gg    \^a-\-x-\-Va-x_a  .^    Va  +  Vx_     2  Vx        (x+a)2 
Va  +  X  -  Va  -  X      X  Va-Vx     y/a  +  Vx    a{x-a) 

Hint.  Rationalize. 

Va  —  X  4-  Vx  —  6         ja  —  x  ^^      ^  ,  y/b-Va_  1       Va—Vb 


41.  V'J^  +  v^^     /«E^.      42.^+; 
Va  -  X  -  Vx-6       \  X  -  6 


V6  Vx         Va 


QUADRATIC  EQUATIONS  77 

^2    2a-(l  +  a2)a;^26  +  (l  +  62)x 

4 


44.  2v^+4-3V2x-3 


Vx  +  4 

45.  (a  -  a;)2  +  (&  _  x)2  =  f  (a  -  cc)  (6  -  x). 

46.  a&x2  -  (a  +  6)  (a6  +  1)  x  +  (a6  +  1)2  =  0. 


47.  V2x-2+V3x  +  7  =  V2x+ll  +  V3x-8. 

48.  V{a  +  x)  (X  +  6)  +  V(a  -  x)  (x  -  b)  =  2  Vox. 


49.  2  72  a  +  6  +  2  X  -  Vl0a  +  6-6x  =  Vl0a  +  96-6x. 

93.  Quadratic  form.    Any  equation  is  in  quadratic  form  if  it 

may  be  written  as  a  trinomial  consisting  of  a  constant  term  and 
two  terms  involving  the  variable  (or  an  expression  which  may  be 
considered  as  the  variable),  the  exponent  in  one  term  being  twice 
that  in  the  other.  By  the  constant  term  is  meant  the  term  not 
containing  the  variable. 

Thus  X  -S  V^  +  13  =  0,  a;-^  +  x"^  -  3  =  0 ,  a2x-2»  _  (a  +  6)  a;-«  +  62  ^  0, 
x2  —  2x  —  S—  Va;2  _  2  x  —  3  + 17  =^  0  are  all  in  quadratic  form.  In  the  last  the 
whole  expression  x2  _  2  x  —  3  is  taken  as  the  variable. 

It  is  usually  convenient  to  replace  by  a  single  letter  the  lower 
power  of  the  variable  or  expression  with  respect  to  which  the 
equation  is  in  quadratic  form. 

EXERCISES 

Solve  and  check  the  following : 

1.  X  -  8  v^  +  15  =  0.  (1) 

Solution :  Let  y/x  =  y. 

Then  x  =  y2. 

Substituting,  (1)  becomes    y2  _  3^  _j.  15  =  0. 
Factor,  (y  -  6)  (y  -  3)  =  0. 

The  roots  are  y  =  5,  y  =  S. 

Thus  Vx  =  6,  Vx  =  3, 

or  X  =  25,  X  =  9. 

Check:  25 -8-5  + 15  =  0;  9-8-3  + 15  =  0. 


78  QUADRATICS  AND  BEYOND 


2.  x-i  - 

5x-§  +  4  = 

=  0. 

Solution : 

Let 

X-3  =  y. 

Then 

X-t  =  y1. 

Substituting,  (1)  becomes 

2/2_  5y +  4  =  0. 

Factor, 

(y  -  4)  (y  -  1)  =  0. 

The  roots  are 

2/ =  4,   y  =  \. 

Since 

x-l-^-     '    . 
x3      ^' 

we  have    ' 

VX2                    ^2 

Thus 

4' 

X2  = 

1                            1 

=  64'     "^-=^8' 

and 

1' 

X2  = 

:1;        x=±Vl  =  ±l. 

(1) 


3.  2Vx2-2x-3  +  x2  -  2x  = 
Solution  :   Add  —  3  to  both  members  and  rearrange  terms, 

x2  _  2x  -  3  +  2  Vx2  -  2x  -  3  -  3  =  0.  (1) 

Let  Vx2  -  2  X  -  3  =  y. 

Then  x2-2x -3  =  2/2. 

Substituting,  (1)  becomes  ^2  ^  2  y  —  3  =  0. 

.     Factor,  (y  +  3)  (y  -  1)  =  0. 

The  roots  are  y  =  —3,  y  =  1. 

Hence  y2  _  ^2  _  2  x  -  3  =  9, 

or  x2_2x  +  l  =  13. 

Extract  the  square  root,  x  —  1  =  ±  "s/l3. 

The  roots  are  x  =  1  ±  VlS. 

Also  x2  -  2  X  -  3  =  1, 

or  x2-2x  +  l  =  5. 

Extract  the  square  root,  x  —  1  =  ±  V5. 

The  roots  are  x  =  1  ±  VB. 

4.  x8  -  1  =  0.  5.  x8  -  8  =  0. 

6.  x8-l  =  0.  7.  x5  +  8xi  =  9x. 

8.  X  -  6  x-i  =  1.  Hint.  Divide  by  Vx.  This  factor  cor- 

responds to  the  root  z  =  0. 

9-  4x3  + 5x^-1  =  0.  10.  10x4-21  =  x2. 


QUADRATIC  EQUATIONS  79 

11.  2x^-3x^-\-x  =  0.  12.  3v^  +  6-l^=4. 

13.  ax^p  +  bxp  +  c  =  0.  14.  3  ic*  -  7  x2  -  6  =  0. 

15.  4x6  -  14x3  +  6  =  0.  16.  x*  -  13x2  +  36  =  0. 

17.  x~  12Vx  +  ll  =  0.  18.  x3  +  4x^=  16ixVx2. 

19.  ■\/x3-2Vx  +  x  =  0.  20.  8X-6  + 999x-3=  125. 

21.  7^x6  +  5xv^  =  66.  22.  2(V5  -  3)=*  -  3  =  v^. 

23.  (x2  -  10)  (x2  -  3)  =  78.  24.  (x2  _  1)2  +  (a;2  +  i)  =  2. 

25.  ( v^  -  1)^  +  v^  =  ^z.  26.  x^  +  x^  =  (28  +  2-8)x V^xl 

27.  (Vx-  3)(v'x  -  4)  =  12.  28.  (X  +  a)^  +  (x  +  6)*  =  (a  -  b)K 

29.  (2x2-3)2-(x2  +  4)2  =  7.  30.  2x2  +  3  Vx2  -  x  +  1  =  2x  +  3. 

31.  (2x  +  3)'  +  (2x  +  3)-^:^0.  32.  8(8x  -  5)8  +  5(5  -  8x)«  =  85. 

33.  x4-4(a+6)x2+16(a-6)2  =  0.  34.  4x2  +  12xVTTi  =  27(1  +  x). 

35.  (2x2-3x+l)2=22x2-33x  +  l.  36.  (x2_  5x+7)2 -(x-2)(x-3)  =  l. 


40.  ^^  =  x-8. 


37.  x2  +  5  =  8x  +  2  Vx2  -  8x  +  40.     38.  2  V(x  -  4)2  +  4(x  -  4)"^  =  9. 

x-4 
2  +  x 
42.  a  =  x2  +  6  +  -. 

44.1+^-     ,       ^' ,   =3. 

2       2(l  +  vT+^)' 

.-      7  ,  41  Vx        97     .     5  ._        2x  (x-l)2       . 

45.  X*  H = h  x5.  46. h  ^^ =  4. 

X  ^^  (x-l)2  2x 

47.  (X  -  1)2  +  4 (X  -  1)  -  5  =  0.  Hint.  I.ety  =  -^^:  then- =  ^^~^^''' 

^  '  ^  '  "       (X  — 1)2  y  2x 

48.   (x2  +  2)^+        ^        =4x2  +  8. 
Vx2  +  2 

94.  Problems  solvable  by  quadratic  equations.  The  principle 
of  translating  the  problem  into  algebraic  symbols,  explained  in 
§  55,  should  be  observed  here.  The  result  should  be  verified  in 
every  case.  It  may  happen  that  the  problem  implies  restrictions 
that  are  not  expressed  in  the  equation  to  which  the  problem 
leads.  In  this  case  some  of  the  solutions  of  the  equation  may 
not  be  consistent  with   the   problem ;    for  instance,   when  the 


80  QUADKATICS  AND  BEYOND 

variable  stands  for  a  nuinber  of  men  fractional  solutions  should 
be  rejected.  If  only  such  results  are  obtained,  the  problem  is 
seK-contradictory.  Often  negative  solutions  should  be  rejected, 
as  when  the  result  indicates  a  negative  number  of  digits  in  a 
number.  Imaginary  or  complex  (p.  72)  results  in  general  mean 
that  the  conditions  of  the  problem  cannot  be  realized. 

PROBLEMS 

1.  The  product  of  ^  and  ^  of  a  certain  number  is  500.  "What  is  the 
number  ? 

2.  There  are  two  numbers  one  of  which  is  less  than  100  by  as  much  as 
the  other  exceeds  it.    Their  product  is  9831.    What  are  the  numbers  ? 

3.  The  sum  of  the  square  roots  of  two  numbers  is  VtI.  One  of  the 
numbers  is  less  than  37  by  as  much  as  the  other  exceeds  it.  What  are  the 
numbers  ? 

4.  A  man  sells  oranges  for  -^^  as  many  cents  apiece  as  he  has  oranges. 
He  sells  out  for  |3.    How  many  had  he  ? 

5.  If  the  perimeter  of  a  square  is  100  feet,  how  long  is  its  diagonal  ? 

6.  A  man  sells  goods  and  makes  as  much  per  cent  as  ^  the  number  of 
dollars  in  the  buying  price.    He  made  $246.    What  was  the  buying  price  ? 

7.  Two  bodies  A  and  B  move  on  the  sides  of  a  right  angle.  A  is  now 
123  feet  from  the  vertex  and  is  moving  away  from  it  at  the  rate  of  239  feet 
per  second.  B  is  239  feet  from  the  vertex  and  moves  toward  it  at  a  rate  of 
123  feet  per  second.    At  what  time  (past  or  future)  are  they  850  feet  apart  ? 

8.  What  is  the  number  twice  whose  square  exceeds  itself  by  190  ? 

9.  What  numbers  have  a  sum  equal  to  63  and  a  product  equal  to  612  ? 

10.  The  sum  of  the  squares  of  two  numbers  whose  difference  is  12  is 
found  to  be  1130.    What  are  the  numbers  ? 

11.  By  what  number  must  one  increase  each  factor  of  24  •  20  so  that  the 
product  shall  be  540  greater  ? 

12.  What  numbers  have  a  quotient  4  and  a  product  900  ? 

13.  Two  numbers  are  in  the  ratio  of  4  :  5.    Increase  each  by  15  and  the 
difference  of  their  squares  is  999.    What  are  the  numbers  ? 

14.  If  4^  is  divided  by  a  certain  number,  the  same  result  is  obtained  as 
if  the  number  had  been  subtracted  from  4^.    What  is  the  number? 

15.  Separate  900  into  two  parts  such  that  the  sum  of  their  reciprocal 
values  is  the  reciprocal  of  221. 


QUADRATIC  EQUATIONS  81 

16.  The  denominator  of  a  fraction  is  greater  by  4  than  the  numerator. 
Decrease  the  numerator  by  3  and  increase  the  denominator  by  the  same, 
and  the  resulting  fraction  is  half  as  great  as  the  original  one.  What  is  the 
original  fraction  ? 

17.  The  numerator  and  denominator  of  a  fraction  are  together  equal 
to  100.  Increase  the  numerator  by  18  and  decrease  the  denominator  by  16, 
and  the  fraction  is  doubled.    What  is  the  fraction  ? 

18.  A  number  consists  of  two  digits  whose  sum  is  10.  Reverse  the  order 
of  the  digits  and  multiply  the  resulting  number  by  the  origingll  one,  and 
the  result  is  2944.    What  is  the  number? 

19.  The  sum  of  two  numbers  is  200.  The  square  root  of  one  increased  by 
the  other  is  44.    What  are  the  numbers  ? 

20.  The  difference  of  two  numbers  is  10,  and  the  difference  of  their  cubes 
is  20630.    What  are  the  numbers  ? 

21.  Around  a  rectangular  flower  bed  which  is  3  yards  by  4  yards  there 
extends  a  border  of  turf  which  is  everywhere  of  equal  breadth  and  whose 
area  is  ten  times  the  area  of  the  bed.    How  wide  is  it  ? 

22.  Two  bicyclists  travel  toward  each  other,  starting  at  the  same  time 
from  places  51  miles  apart.  One  goes  at  the  rate  of  9  miles  an  hour.  The 
number  of  miles  per  hour  gone  by  the  other  is  greater  by  5  than  the  number 
of  hours  before  they  meet.     How  far  does  each  travel  before  they  meet  ? 

23.  A  printed  page  has  15  lines  more  than  the  average  number  of  letters 
per  line.  If  the  number  of  lines  is  increased  by  15,  the  number  of  letters  per 
line  must  be  decreased  by  10  in  order  that  the  amount  of  matter  on  the  two 
pages  may  be  the  same.    How  many  letters  are  there  on  the  page  ? 

24.  A  merchant  buys  goods  for  a  certain  sum.  The  cost  of  handling  them 
was  5%  of  their  cost  price.  He  sells  for  §504,  gaining  as  much  per  cent  as 
T^Q  the  cost  price  was  in  dollars.    What  was  the  cost  price  ? 

25.  A  man  had  $8000  at  interest.  He  increased  his  capital  by  $100  at  the 
end  of  each  year,  apart  from  his  interest.  At  the  beginning  of  the  third  year 
he  had  $8982. 80.    What  per  cent  interest  did  his  money  draw  ? 

26.  Two  men  A  and  B  can  dig  a  trench  in  20  days.  It  would  take  A  9 
days  longer  to  dig  it  alone  than  it  would  B.  How  long  would  it  take  B 
alone  ? 

27.  A  cistern  is  emptied  by  two  pipes  in  6  hours.  How  long  would  it 
take  each  pipe  to  do  the  work  if  the  first  can  do  it  in  5  hours  less  time  than 
the  second  ? 

28.  A  party  procures  lunch  at  a  restaurant  for  $16.  If  there  had  been 
5  less  in  the  party,  each  member  would  have  paid  15  cents  more  without 
affecting  the  amount  of  the  entire  bill.    How  many  were  in  the  party  ? 


82  QUADRATICS  AND  BEYOND 

29.  A  party  pays  $12  for  accommodations.  Had  there  been  4  more  in  the 
party,  and  if  each  person  had  paid  25  cents  less,  the  bill  would  have  been 
$16.    How  many  were  in  the  party  ? 

30.  A  grocer  sells  his  stock  of  butter  for  $16.  If  he  had  had  6  pounds  less 
in  stock,  he  would  have  been  obliged  to  charge  10  cents  more  a  pound  to 
realize  the  same  amount.    How  many  pounds  had  he  in  stock  ? 

31.  A  man  buys  lemons  for  $2.  If  he  had  received  for  that  money  50 
more  lemons,  they  would  have  cost  him  2  cents  less  apiece.  What  was  the 
price  of  each  lemon  ? 

32.  It  took  a  number  of  men  as  many  days  to  dig  a  ditch  as  there  were 
men.  If  there  had  been  6  more  men,  the  work  would  have  been  done  in  8 
days.    How  many  men  were  there  ? 

95.  Theorems  regarding  quadratic  equations.  In  this  and  the 
following  sections  we  prove  several  theorems  concerning  quad- 
ratic equations.  Similar  theorems  are  later  proved  in  general  for 
equations  of  higher  degree. 

Theorem.    If  a  is  a  root  of  the  equation 

ac(^-\-hx-\-c  =  0,  (1) 

then  X  —  a  is  a  factor  of  its  left-hand  member ,  and  conversely. 

The  fact  that  a  is  a  root  of  the  equation  is  equivalent  to  the 
assertion  of  the  truth  of  the  identity 

aa^  -\-  ha  ■\-  G  =i  Oj 

by  definition  of  the  root  of  an  equation  (§  53). 
Divide  ax^  4-  Jx  -f  c  by  ic  —  a  as  follows : 

X  —  a\ax^  -\-  bx  -\-  c\ax  +  (^  4-  ao^) 


(b  +  aa)  X  -{-  c 

(b  -f-  aa)  X  —  aa^  —  ba 

aa^  -{-  ba  -\-  c 

Since  the  remainder  vanishes  by  hypothesis,  ax^  +  bx  -{-  c  is 
exactly  divisible  hy  x  —  a. 

Conversely,  we  have  already  seen  (p.  3)  that  ii  x  —  a  is  a, 
factor  of  an  equation,  a  is  a  root,  since  replacing  x  hy  a  would 
make  that  factor  vanish. 


QUADRATIC  EQUATIONS  83 

EXERCISES 

Form  equations  of  which  the  following  are  roots. 

1.  2,  6. 

Solution :  Since  2  and  6  are  roots,  (x  —  2)  and  (x  —  6)  are  factors,  and  the 
equation  having  these  as  factors  is 

(X  -  2)  (X  -  6)  =  x2  -  8  X  +  12  =  0. 

2.  1,  -  1.  3.-3,-4. 
4.   V2,  -  V2.  5.   -  2,  -  6. 

6.   -  4  V2,  V32.  7.   V27,  -  3  VS. 

8.  2  +  V3,  2  -  V3.  9.  a  +  V6,  a  -  V6. 

96.  Theorem.    A  quadratic  equation  has  only  two  roots. 

Given  the  equation  ax^  +  ?)X  +  c  =  0  with  the  roots  a  and  /3,  to  prove  that 
the  equation  has  no  other  root,  as  7,  distinct  from  a  and  /3. 

Since  a  and  j8  are  roots  of  the  equation,  x  —  a  and  x  —  /3  are  factors. 
Thus  our  equation  may  be  written  in  the  form 

a(x-a:)(x-/3)  =  0.  (1) 

If  now  7  is  a  root,  it  must  satisfy  (1),  i.e. 

a(7-a)(7-/3)  =  0. 

But  in  order  that  any  product  of  numerical  factors  should  vanish  one  of 
the  factors  must  vanish.  Tlius  either  a  =  0,  or  7  —  or  =  0,  or  7-/3  =  0. 
But,  by  hypothesis,  7  5^  <x  and  7  ^  /S,  so  the  last  two  factors  cannot  vanish. 
Thus  a  =  0.  This  would,  however,  reduce  our  equation  to  a  linear  equa- 
tion, which  is  contrary  to  our  hypothesis  that  the  equation  is  quadratic. 
Thus  the  assumption  that  we  have  a  third  distinct  root  leads  to  a  contradiction. 

CoKOLLARv  I.  If  a  quadratic  equation  is  satisfied  by  more  than  two  distinct 
values  of  the  variable,  then  each  of  the  coefficients  vanishes  identically. 

The  above  proof  shows  that  the  coefficient  of  x^  must  vanish.  In  the  same 
way  it  can  be  shown  that  b  =  c  =  0. 

Corollary  II.  If  two  quadratic  expressions  have  the  same  value  for  more 
than  two  values  of  the  variable,  then  their  coefficients  are  identical. 

Let  ax2  +  6x  +  c  =  a'x^  +  b'x  +  c' 

for  more  than  two  values  of  x.    Transpose,  and  we  obtain 
(a  -  a')x^  +  (6  -  b')x  +  c  -  c'  =  0. 

We  have  then  a  quadratic  equation  satisfied  by  more  than  two  values 
of  X.  Thus  by  Corollary  I  each  of  its  coefficients  must  vanish.  Thus  a'  =  a, 
6'  =  6,  c'  =  c. 


84  QUADRATICS  AND  BEYOND 

This  theorem  taken  with  §  95  is  equivalent  to  the  statement  that  a  quad- 
ratic equation  can  be  factored  in  one  and  only  one  way.    Thus  if 

ax"^  +  6x  4-  c  =  a  (x  —  or)  (x  —  /3), 
we  cannot  find  other  numbers  7  and  5  distinct  from  a  and  /3  such  that 

ax2  +  6x  +  c  =  a  (x  —  7)  (x  —  5), 
for  then  the  equation  would  have  roots  distinct  from  a  and  /3. 

97.  Theorem.    If  the  equation 

x2  +  6x  +  c  =  0,  (1) 

where  b  and  c  are  integers,  has  rational  roots,  those  roots  must  be  integers. 

P 
For  suppose  -  to  be  a  rational  fraction  reduced  to  its  lowest  terms  and 

a  root  of  (1). 


Then 

^,  +  ^-^  +  0  =  0. 
q2       q 

or 

p2  +  ipq  +  c^2  =  0, 

which  gives 

p'^  =  -  q{bp-{-  cq). 

Thus  some 

factor  of  q  must  be  contained  in  p  (§  69),  which  contradicts 

the  hypothesis 

that  the  fraction  -  is  already  reduced  to  its  lowest  terms. 

98.  Nature  of  the  roots  of  a  quadratic  equation. 

The  equation 

ax^  -^  bx -\-  c  =  0 

(1) 

has  as  roots  (§  89)      x^  =         "^  ^  ^^,  (2) 


_  ^  -l_  V^'^  - 

-  4c  ac 

2a 

-b-^b^- 

-  4:  ao 

2a  ■  (^) 

These  expressions  afford  an  immediate  arithmetic  means  of 
determining  the  nature  of  the  roots  of  the  given  equation  when 
a  J  b,  and  c  have,  numerical  values  and  a  =^  0.  In  fact  an  inspec- 
tion of  the  value  of  6^  —  4  ac  is  sufficient  to  determine  the  nature 
of  the  roots  of  (1).  'mM 

I.  When  h^  —  Jf.ac  is  negative,  the  roots  are  iikhg%naty  (§89). 
II.  When  y^  —  Ji^ac^Oy  the  roots  are  real  and  equal.    In  this 

b 
case  x^=x^=-—- 

III.  When  y^  —  Ji^ac  is  positive,  the  roots  are  real  and  distinct. 

IV.  When  ¥  —  Jiac  is  positive  and  a  perfect  square,  the  roots 
are  real,  distinct,  and  rational. 


QUADRATIC  EQUATIONS  85 

In  case  IV,  if  J^  —  4  ac  =  A, 

_  —  h  +  Va  _  —  ^,  -  Va 

^'  ~         2  a       '   ^"^  "         2  a        ' 

The  converses  of  these  four  cases  are  also  true.  For  instance, 
if  the  roots  of  (1)  are  imaginary,  from  (2)  and  (3)  it  is  clear 
that  P  —  4:  ac  must  be  negative. 

The  expression  A  =  Z>^  —  4  ac  is  called  the  discriminant  of  the 
equation  ax^  -\-  bx  -^  c  =  0. 

EXERCISES 

1.  Determine  the  nature  of  the  roots  of  the  following  equations  without 
solving. 

(a)  3x2  _4x-l  =0. 

Solution  :  A  =  (-  4)2  -  4  .  3  .  (-  1)  =  16  +  12  =  28  and  is  then  positive. 
Thus  by  III  the  roots  are  real  and  distinct. 

(b)  3x2  -  7x  +  6  =  0.  (c)  6x2  -  X  -  1  =  0. 
(d)  3x2  +  4x  +  l  =  0.  (e)  x2-4x  +  l  =  0. 
(f)  2x2 -6x-9=:0.  (g)  2x2-4x-2  =  0. 
(h)  4x2  +  12x  +  9  =  0.  (i)  2x2  +  6x  -  4  =  0. 

(j)  4x2- 28x  + 49  =  0.  (k)  4x2  +  12x  + 5  =  0. 

2.  Determine  real  values  of  k  so  that  the  roots  of  the  following  equations 
may  be  equal.    Check  the  result. 

(a)  (2  + A;)x2  +  2A:x  +  l  =  0. 

Solution :  Here  2  -\-  k  =  a,       2fc  =  6,         l  =  c. 

Thus  A  =  62-4ac  =  4fc2_4.(fc  +  2).l 

=  4  ^2  _  4  A:  _  8. 
Since  the  roots  of  an  equation  are  equal  when  and  only  when  its  dis- 
criminant equals  zero  (§  98,  II),  the  required  values  of  k  make  A  =  0  and 

are  the  roots  of 

4^2  _  4^  _  8  =  0, 

or  fc2  -  A:  -  2  =  0. 

Solve  by  factoring, 

k^-k-2=:{k-2){k  +  l)  =  0. 
Thus  the  values  of  k  are  /c  =  2,     k  =  —  1. 

Check :  Substituting  in  the  original  equation  for  /c  =  2,  we  get 
4x2  +  4x  +  l  =  (2x  +  l)2, 
and  for  A:  =  —  1  we  get        x2  —  2  x  +  1  =  (x  —  1)2. 


86  QUADRATICS  AND  BEYOND 

(b)  x2  +  fcx  +  16  =  0.  (c)  a;2  +  2a;  +  A;2  =  0. 

(d)  x2-2A:x  +  l  =  0.  (e)  3A;x2-4x-2  =  0. 

(f)  A;x2-3x  +  4  =  0.  (g)  x^ -\- 4kx -\- k^ +  1  =  0. 

(h)  A:2x2  4-  3x  -  2  =  0.  (i)  (A:2  +  3) x2  +  fee  -  4  =  0. 

(j)  3/cx2  +  A-x  -  1  =  0.  (k)  x2  +  (3A;  +  l)x  +  1  =  0. 

(1)  x2  +  3x  +  A;  -  1  =  0.  (m)  x2  +  9A;x  +  6fc  +  ^  =  0. 

(n)  4 A;2x2  +  4  A-x  -  125  =  0.  (o)  2  x2  -  4x  -  2  A;  +  3  ==  0. 

(p)  (A:  +  l)x2  +  A:x  +  A;  +  2  =  0.  (q)  A;x2  +  (4  A;  +  l)x  +  4 A:  -  3  =  0. 

(r)  2(A:4- l)x2  +  3A:x  + A;-1  =  0.  (s)  (A:  -  l)x2  +  5A:x  +  6A;  +  4  =  0. 

(t)  (2A;  +  3)x2-7A;x+^^^  =  0.  (u)  (A;-l)x2+(2A; +l)x+ A:  +  3  =  0. 


CHAPTEE  IX 
GRAPHICAL  REPRESENTATION 

99.  Representation  of  points  on  a  line.  Let  us  select  on  the 
indefinite  straight  line  AB  sl  certain  point  0  as  a  point  of  refer- 
ence. Let  us  also  select  a  certain  line,  the  length  of  which  for 
the  purpose  in  hand  shall  represent  unity.  Let  us  further  agree 
that  positive  numbers  shall  be  represented  on  ^J5  by  points  to 
the  right  of  0,  whose  distances  from  0  are  measured  by  the  given 

-3       -2       -1  +1      +2+3  0  1 


numbers,  and  negative  numbers  similarly  by  points  to  the  left. 
Then  there  are  certainly  on  AB  points  which  represent  such  num- 
bers as  2,  —  3,  ^,  —  1^,  or,  in  fact,  any  rational  numbers.  Since 
we  can  divide  a  line  into  any  desired  number  of  equal  parts,  we 
are  able  to  find  by  geometrical  construction  the  point  correspond- 
ing to  any  rational  number.  Furthermore,  by  the  principle  that 
the  square  of  the  hypotenuse  of  a  right  triangle  equals  the  sum 
of  the  squares  of  the  other  two  sides,  we  can  find  the  point 
corresponding  to  any  irrational  number  expressed  by  square-root 
signs  over  rational  numbers.  More  complicated  irrational  num- 
bers cannot,  however,  in  general  be  constructed  by  means  of  ruler 
and  compasses,  but  we  assutne  that  to  every  real  number  there 
corresponds  a  point  on  the  line,  and  conversely ,  we  assert  that  to 
every  point  on  the  line  corresponds  a  real  number.  This  assump- 
tion of  a  one-to-one  correspondence  between  points  and  real  num- 
bers is  the  basis  of  the  graphical  representation  of  algebraic 
equations. 

This  amounts  to  nothing  more  than  the  assertion  that  every  real  number, 
rational  or  irrational,  as,  for  instance,  —  6,  2  +  V3,  V3,  tt,  represents  a  certain 
distance  from  0  on  AB,  and  conversely,  that  whatever  point  on  the  line  we  may 
select,  the  distance  from  0  to  that  point  may  he  expressed  by  a  real  number. . 

87 


88 


QUADRATICS  AND  BEYOND 


'<Y 


X  axis 


T^ 


100.  Cartesian  coordinates.  We  have  seen,  that  when  the 
single  letter  x  takes  on  real  values  all  these  values  may  be  repre- 
sented by  points  on  a  straight  line. 
When,  however,  we  have  two  variables, 
as  X  and  y,  which  we  wish  to  represent 
simultaneously,  we  make  use  of  the 
plane.  Just  as  we  determined  arbitra- 
rily, on  the  line  along  which  the  single 
variable  was  represented,  an  arbitrary 
point  for  the  point  of  reference  and  an 
arbitrary  length  for  the  unit  distance, 
so  now  we  select  an  indefinite  line  along  which  x  shall  be  repre- 
sented, and  another  perpendicular  to  it  along  which  7/  shall  be 
represented.  The  former  we  call  the  X  axis ;  the  latter  the  Y  axis. 
The  intersection  0  of  these  axes  we  take  as  the  point  of  reference 
for  each.    This  point  is  called  the  origin. 

We  select  a  unit  of  distance  for  x  and  a  unit  of  distance  for  1/ 
-v^hich  may  or  may  not  be  the  same,  according  to  the  problem 
under  discussion.  As  before,  we  represent  positive  numbers  on 
the  X  axis  to  the  right,  and  negative  numbers  to  the  left.  Positive 
values  of  y  are  represented  above  the  X  axis,  and  negative  values 
below  it.  The  arrowhead  on  the 
axes  indicates  the  positive  direc- 
tion. Any  pair  of  values  of  x 
and  y,  written  (x,  ?/),  may  now 
be  represented  by  a  point  on 
the  plane  which  is  x  units  from 
the  Y  axis  and  y  units  from  the 
X  axis.  Thus  if  x  =  0,  y  =  0, 
written  (0,  0),  the  point  repre- 
sented is  the  origin.  The  point 
(3,  0),  i.e.  a;  =  3,  ?/  =  0,  is  found 
by  going  three  units  of  x  to  the 
right,  i.e.  in  the  positive  direction  of  x  and  no  units  up.  The  point 
(4,  3),  i.e.  a  =  4,  2/  =  3,  is  found  by  going  four  units  of  x  to  the 
right  and  three  units  of  y  up.    The  point  (—  3,  1)  is  found  by 


3,4) 


YA 


0(0,7) 


U',3. 


3.0) 


{4, -2) 


GRAPHICAL  REPRESENTATION  89 

going  three  units  of  x  to  the  left  and  one  unit  of  y  up.  The  point 
(—  3,  —  4)  is  found  by  going  three  units  of  x  to  the  left  and  four 
units  of  y  down.  In  fact,  if  we  let  both  x  and  y  take  on  every 
possible  pair  of  real  values,  we  have  a  point  of  our  plane  corre- 
sponding to  each  pair  of  values  of  (x,  y).  Conversely,  to  every 
point  of  the  plane  correspond  a  pair  of  values  of  (x,  y).  These 
values  are  called  the  coordinates  of  the  point.  The  value  of  a?, 
i.e.  the  distance  of  the  point  from  the  Y  axis,  is  called  its 
abscissa;  the  value  of  y,  i.e.  the  distance  of  the  point  from  the 
X  axis,  is  called  its  ordinate.  If  the  point  (x,  y)  is  conceived  as  a 
moving  point,  and  if  no  restriction  is  placed  . 

on  the  value  of  the  coordinates  so  that  they 

^  .,  T  'PIT  ^>^<^  Quadrant     ist  Quadrant 

take  on  every  possible  pair  oi  real  values,       (-,+;        (-|-,+) 


ith  Quadrant 


every  point  in  the  plane  is  reached  by  the     ^ 

moving    point     (X,    y).  3rd  Quadrant 

The  X  and  Y  axes  divide  the  plane  into       (-'-) 
four  parts  called  quadrants,  which  are  num- 
bered as  in  the  figure.    The  proper  signs  of  the  coordinates  of 
points  in  each  of  the  quadrants  are  also  indicated. 

EXERCISES 

The  following  exercises  should  be  carefully  worked  on  plotting  paper, 
which  can  be  bought  ruled  for  the  purpose. 

1.  Plot  the  points  (2,  3),  (0,  4),  (-  4,  0),  (-  9,  -  2),  (2,  -  4). 

2.  Plot  with  the  aid  of  compasses  the  points  (l,  V'2),  (V3,  —  V2), 
(2+V3,  2-V3),(-V2,  -V2). 

3.  Plot  the  square  three  of  whose  vertices  are  at  (—  1,  —  1),  (—  1,+  1), 
(+  1,  —  1).    What  are  the  coordinates  of  the  fourth  vertex? 

4.  Plot  the  triangle  whose  vertices  are  (2,  1),  (—6,  —  2),  (—4,  4). 

5.  Plot  the  two  equilateral  triangles  two  of  whose  vertices  are  (6,  1), 
(—6,  1).    Find  coordinates  of  the  remaining  vertices. 

6.  If  the  values  of  the  coordinates  (x,  y)  of  a  moving  point  are  restricted 
so  that  both  are  positive  and  not  equal  to  zero,  where  is  the  point  still  free 
to  move  ? 

7.  If  the  coordinates  {x,  y)  of  a  moving  point  are  restricted  so  that  con- 
tinually y  =  0,  where  is  the  point  still  free  to  move  ? 


90 


QUADRATICS  AND  BEYOND 


8.  What  is  the  abscissa  of  any  point  on  the  T  axis  ? 

9.  The  coordinates  of  a  variable  point  are  restricted  so  that  its  ordinate 
is  always  2.    Where  may  the  point  move  ? 

10.  If  both  ordinate  and  abscissa  of  a  point  vanish,  can  the  point  move  ? 
Where  will  it  be  ? 

11.  Plot  the  quadrilateral  whose  vertices  are  (0,  0),  (—  6,  —  3),  (5,  —  5), 
(—1,  —  8).    What  kind  of  a  quadrilateral  is  it? 

12.  The  coordinates  of  three  vertices  of  a  parallelogram  are  (—  1,  —  1), 
(6,  2),  (—1,  —  6).    Find  the  coordinates  of  the  fourth  vertex. 

13.  The  coordinates  of  two  adjacent  vertices  of  a  square  are  ( —  1,  —  2)  and 
(1,  —  2).  Find  the  coordinates  of  the  remaining  vertices  (two  solutions).  Plot 
the  figures. 

14.  The  coordinates  of  two  adjacent  vertices  of  a  rectangle  are  (—  1,  —  2), 

(1,  —  2).     What  restriction  is  imposed  on  the   coordinates  of   remaining 
vertices  ? 

15.  The  coordinates  of  the  extremities  of  the  bases  of  an  isosceles  triangle 
are  (1,  6),  (1,  —  2).  Where  may  the  vertex  lie?  What  restriction  is  imposed 
on  the  coordinates  (x,  y)  of  the  vertex  ? 

101.  The  graph  of  an  equation.  The  equation  a?  =  2  ?/  is  satis- 
fied by  numberless  pairs  of  values  (x,  ?/);  for  example,  (2,  1), 
(0,  0),  (1,  ^),  (—  2,  —  1)  all  satisfy  the  equation.  There  are,  how- 
ever, numberless  pairs  of  values  which  do  not  satisfy  the  equation ; 

for  example,  (1,  2),  (2,  - 1),  (- 1, 1), 
(0,  —  1).  The  pairs  of  values  which 
satisfy  the  equation  may  be  taken 
as  the  coordinates  of  points  in  a 
plane.  The  totality  of  such  points 
would  thus  in  a  sense  represent  the 
equation,  for  it  would  serve  to  dis- 
tinguish the  points  whose  coordi- 
nates do  satisfy  the  equation  from 
those  whose  coordinates  do  not. 
After  finding  a  few  pairs  of  values  which  satisfy  the  above  equa- 
tion we  note  that  any  point  whose  abscissa  is  twice  the  ordinate, 
i.e.  for  which  cc  =  2  ?/,  is  a  point  whose  coordinates  satisfy  the 
equation.  Any  such  point  lies  on  the  straight  line  through  the 
origin  and  the  point  (2,  1).    We  can  then  say  that  those  points 


I 


GRAPHICAL  REPRESENTATION  91 

and  only  those  which  are  on  the  straight  line  represented  in  the 
figure  have  coordinates  which  satisfy  the  equation.  This  line  is 
the  graphical  representation  or  graph  of  the  equation. 

The  equation  of  a  line  or  a  curve  is  satisfied  by  the  coordinates 
of  every  point  on  that  line  or  curve. 

Any  point  whose  coordinates  satisfy  an  equation  is  on  the 
graph  of  the  equation. 

102.  Restriction  to  coordinates.  Iri  §  100  it  was  seen  that  a 
moving  point  whose  coordinates  were  unrestricted  took  on  every 
position  in  the  plane.  We  now  see  that  when  the  coordinates  of 
a  point  are  restricted  so  as  to  satisfy  a  certain  equation  (as  x  =  2  3/), 
the  motion  of  the  point  is  no  longer  free,  but  restrained  to  move 
along  a  certain  path.  Thus,  for  instance,  the  equation  a;  =  4  means 
that  the  path  of  the  moving  point  is  so  restricted  that  its  abscissa 
is  always  4.  Its  ordinate  is  still  unrestricted  and  may  have  any 
value.  This  shows  that  the  plot  of  a?  =  4  is  a  straight  line  four 
units  to  the  right  of  the  Y  axis  and  parallel  to  it,  for  the  abscissa 
of  every  point  on  that  line  is  4,  and  every  point  whose  abscissa  is 
4  lies  on  that  line. 

EXERCISES 

Determine  on  what  line  the  moving  point  is  restricted  to  move  by  the 
following  equations.    Draw  the  graph. 

1.  x  =  6.  2.  x  =  0.  3.  ?/  =  f. 

4.  X  =  y.  h.  y  =  2.  6.  3  X  =  y. 


7.  2x  =  y.  8.  y  =  0.  9.  x  =  -3. 

10.  X  =  32/.  11.  3y  =  -  X.  12.  X  +  y  =  0. 

13.  6x  =  ll.  14.  2/=-3.  15.  2x  =  -3?/. 

16.  x  =  -2y.  17.  x  =  -l.  18.  2x-62/  =  0. 

103.  Plotting  equations.  Plotting  an  equation  consists  in  find- 
ing the  line  or  curve  the  coordinates  of  whose  points  satisfy  the 
equation.  Thus  the  process  of  §  101  was  nothing  else  than  plot- 
ting the  equation  x  =  2y.  This  may  be  done  in  some  cases  by 
observing  what  restriction  the  equation  imposes  on  the  coordinates 
of  the  moving  point ;  but  more  often  we  are  obliged  to  form  a 


92 


QUADRATICS  AND  BEYOND 


table  of  various  solutions  of  the  equation,  and  to  form  a  curve 
by  joining  the  points  corresponding  to  these  solutions.  This 
gives  us  merely  an  approximate  figure  of  the  exact  graph  which 
becomes  more  accurate  as  we  find  the  coordinates  of  points  closer 
to  each  other  on  the  line  or  curve. 

EuLE.  Wlien  y  is  alone  on  one  side  of  the  equation,  set  x  equal 
to  convenient  integers  and  compute  the  corresponding  values  of  y. 

Arrange  the  results  in  tabular  form.  Take  corresponding 
values  of  x  and  y  as  coordinates  and  plot  the  various  points. 

Join  adjacent  points,  making  the  entire  plot  a  smooth  curve. 

When  X  is  alone  on  one  side  of  the  equation  integral  values  of  y  may  be  assumed 
and  the  corresponding  values  of  x  computed. 

Care  should  be  taken  to  join  the  points  in  the  proper  order  so  that  the  resulting 
curve  pictures  the  variation  of  y  when  x  increases  continuously  through  the  values 
assumed  for  it.  By  adjacent  points  we  mean  points  corresponding  to  adjacent 
values  of  x. 

Any  scale  of  units  along  the  X  and  Y  axes  that  is  convenient  may  be  adopted. 
The  scales  should  be  so  chosen  that  the  portion  of  the  curve  that  shows  considerable 
curvature  may  be  displayed  in  its  relation  to  the  axes  and  the  origin. 

When  there  is  any  question  regarding  the  position  of  the  curve  between  two 
integral  values  of  x,  an  intermediate  fractional  value  of  x  may  be  substituted,  the 
corresponding  value  of  y  found,  and  thus  an  additional  point  obtained  to  fix  the 
position  of  the  curve  in  the  vicinity  in  question. 


EXERCISES 
Plot: 

1.  x2  -4a;  +  3  =  y- 

Solution :  In  this  equation  if  we  set  x  =  0, 
1,  2,  3,  etc.,  we  get  3,  0,  —  1,  0,  etc.,  as  cor- 
responding values  of  y.  Thus  the  points 
(0,  3),  (1,  0),  (2,  -  1),  (3,  0),  etc.,  are  on 
the  curve.  These  points  are  joined  in  order 
by  a  smooth  curve. 


X 

y 

X 

y 

0 

3 

-1 

8 

1 

0 

-2 

15 

2 

-1 

3 

0 

4 

3 

6 

8 

6 

16 

\ 

W^ 

1/ 

1 

i 

\ 

/ 

\ 

/ 

\ 

/ 

' 

/■ 

i 

\ 

/ 

\ 

f 

0 

\ 

/ 

*x 

— 

— 

— 

- 

GRAPHICAL  REPRESENTATION 


93 


2.  y  =  x^-lx  +  l. 
4.  y  =  x^  -  Sx  +  2. 
6.  y  =  x^  —  2x  +  1. 
8.  2/  =  2x2-6x  +  7. 
10.  y  =  2x^-6x-3. 


3.  y  =  x2  +  1. 

5.  y  =  x^  —  4x. 

7.  y  =  x'^  +  6x  +  5, 

9.  y  =  2x2^-3x  +  4. 

11.  2/  =  x2-12x  +  ll. 


104.  Plotting  equations  after  solution.  When  neither  x  nor  y 
is  abeacly  alone  on  one  side  of  the  equation,  the  equation  should 
be  solved  for  y  (or  x)  and  the  rule  of  the  previous  section  applied. 
It  should  be  noted  that  when  a  root  is  extracted  two  values  of 
y  may  correspond  to  a  single  value  of  x. 


EXERCISES 


Plot: 

1.  2x2+  3y2  =  9. 

Solution 


32/2  =  9 
2/2  =  3 


2x2, 


3/  =  ±  V3-fx2. 
Assuming  the  various  integral  values  for  x,  we  obtain  the  following 


table  and  plot: 


X 

y 

0 

±V3  =  ±1.7 

1 

±VI  =  ±i-5 

2 

±Vi=±    -67 

+  3 

imaginary 

X 

y 

-1 
-2 
-3 

±VJ  =  ±    -57 
imaginary 

N: 


In  this  example,  when  x  is  greater  than  3  or  less  than  —  S,  y  is  imaginary. 
Thus  none  of  the  curves  is  found  outside  a  strip  x  =  ±S. 

To  find  exactly  where  the  curve  crosses  the  X  axis,  the  equation  may  be 
solved  for  x  and  the  value  of  x  corresponding  to  y  =  0  found.    Thus 


X  =  ±  VF^I^'- 

If  2/  =  0,  X  =  ±  Vl  =  2-1.    This  point  is  included  in  the  plot 
2.  X2/  =  4. 


Solution : 


4 


94 


QUADRATICS  AND  BEYOND 


Form  table  for  integral  values  of  x. 


i 


X 

y 

-1 

-4 

-2 

-2 

-3 

-f 

-4 

-1 

-6 

-1 

-8 

-\ 

-12 

-i 

Since  this  table  does  not  give  us  any 
idea  of  the  curves  between  -\- 1  and  —  1, 
v^e  supplement  the  table  by  assuming 
fractional  values  for  x. 


3.  x^  =  y\ 
5.  xy  =  —  1. 

7.  x2  +  ?/2  =  16. 

9.  a;2 +  2/2  =  25. 

11.  2x2/  +  3x  =  2. 

13.  xy  +  2/2  =  10. 

10  —  2/2 


Hint,  a;  = 


y 


4.  X2/  =  16. 

6.  X2/  =  X  +  1. 

8.  x2  -  2/2  =  9. 

10.  x2  +  X  =  12  2/. 

12.  x2  +  9  2/2  =  36. 

14.  X  —  2/  +  2  xy  =  0. 

2/ 

Hint.  x  = — -• 

1  +  22/ 


2/ 

X 

6 

-1 

8 

-1 

12 

-i 

y 


-8 
12 


15.  6x2+2x  +  32/2  =  0.     16.  x2  +  2x  +  1  =  y2  _  3^. 

105.  Graph  of  the  linear  equation.  The  intimate  relation 
between  the  simplest  equations  and  the  simplest  curves  is  given 
in  the  following  theorems. 

Theorem  I.  The  graph  of  the  equation  y  =  ax  is  the  straight 
line  through  the  origin  and  the  point  (1,  a),  where  a  is  any  real 
number. 

The  proof  falls  into  two  parts. 

First.  Any  point  on  the  line  through  the  origin  and  the  point 
(1,  a)  has  coordinates  that  satisfy  the  equation.  Let  P  (Figure  1) 
with  coordinates  (x',  y')  be  on  the  line  OA.    By  similar  triangles 


V 


or 


ax' 


GRAPHICAL  KEPRESENTATION 


95 


Thus  the  coordinates  of  any  point  on  the  line  satisfy  the  equation. 


C«'y) 


Figure  1 


Figure  2 


Second.  Any  point  whose  coordinates  satisfy  the  equation  lies 
on  the  line. 

Let  the  coordinates  (x\  y')  of  the  point  P  (Figure  2)  satisfy  the 
equation.    Then  we  have 


or 


y  =  ax', 

^,  =  a. 

x' 


Let  the  ordinate  y'  cut  the  line  at  B.    Then  by  the  first  part 
of  the  proof  BC  =  ax', 

EC 


or 


x' 


a. 


Thus 

Hence  P  lies  on  the  line. 


v'      EC 
a=—,  =  — TJ    or  ti'  =  EC. 
x'       x'  -^ 


Theorem  II.    The  graph  of  any  linear  equation  in  two  vari- 
ables is  a  straight  line. 

The  general  linear  equation 

Ax-{-By-{-C  =  0  (1) 

may  be  written  in  the  form 

y  =  ax-\-b,  (2) 

A  C 

where  a  =  —  —  and  b=  —  —f  provided  ^  ^  0  (§  7).     It  E  =  0, 
B  E 

the  equation  Ax  -\-  C  =  0  may  be  put  in  the  form 

C 


96 


QUADRATICS  AND  BEYOND 


provided  A  =^  0.  This  is  evidently  the  equation  of  a  straight 
line  parallel  to  the  Y  axis  (§  102).  li  B  =  0  and  A  =  0,-we  have 
no  term  left  involving  the  variable.    Thus  the  only  case  for 

which  the  theorem  demands  proof  is 
when  B  ^  0,  and  the  equation  may  be 
reduced  to  form  (2).  By  Theorem  I 
we  know  that  the  graph  oi  y  =  ax  is 
a  straight  line.  If,  then,  we  add  to 
every  ordinate  y  of  the  line  y  =  ax  the 
constant  h,  the  locus  of  the  extremi- 
ties of  the  lengthened  ordinates  will 
lie  in  a  straight  line,  as  one  can  easily 
prove  by  Geometry.  But  any  point  (cc,  y)  on  the  upper  line  is 
such  that  its  ordinate  y  is  equal  to  the  ordinate  of  the  lower 
line,  i.e.  ax,  and  in  addition  the  constant  h ;  that  is,  y  =  ax  -\-  h. 
Also,  since  the  upper  line  is  the  locus  of  the  extremities  of  the 
lengthened  ordinates,  every  point  whose  coordinates  satisfy  the 
equation  y  =  ax  +  h  i^  on  this  upper  line.  Thus  the  equation  (1) 
has  a  straight  line  as  its  graph. 

CoEOLLARY.    Two  lines  vjhose  equations  are  in  the  form 

y  =  ax-\-h,  (3) 

yz=ax-\-V  (4) 

are  'parallel.  ^ 

For  the  value  of  the  ordinates  of  (3)  corresponding  to  a  given 
abscissa,  say  x^^  is  obtained  from  the  ordinate  of  (4)  corresponding 
to  the  same  abscissa  by  adding  the  constant  h  —  h\  Thus  each 
point  on  (3)  is  always  found  h  —  V  units  above  (below  if  h  —  b' 
is  negative)  a  point  of  (4).    Thus  the  lines  are  parallel. 

106.  Method  of  plotting  a  line  from  its  equation.  Since  the 
equation  y  =  ax-\-h\^  satisfied  by  the  values  (0,  h),  the  graph  cuts 
the  Y  axis  h  units  above  (below  if  b  is  negative)  the  origin.  Since 
it  is  satisfied  by  the  values  (1,  a  -\-b),  the  graph  passes  through 
the  point  reached  by  going  one  unit  of  x  to  the  right  of  (0,  b)  and 
a  units  up  (down  if  a  is  negative).  These  two  points  determine 
the  line.    We  may  then  plot  a  linear  equation  by  the  following 


GRAPHICAL  REPRESENTATION 


97 


Rule.    Reduce  the  given  equation  to  the  form 
y  —  ax-\-h. 

Plot  the  point  (0,  h)  as  one  of  the  two  points  that  determine 
the  line. 

From  this  point  go  one  unit  of  x  to  the  right  and  a  units 
of  y  up  {down  if  a  is  negative)  to  find  a  second  point  that  lies 
on  the  line. 

Draw  the  line  through  these  two  points. 


EXERCISES 


Plot: 

1.  6x  +  2y-5  =  0. 

Write  in  the  form 

y  =  ax  +  h, 

and  we  have 

y=_3x  +  f. 

a  =-3; 

6  =  f. 

Plot  the  point  (0,  |). 

From  this  point  go  one  unit  of  x  to  the  right  and  three 
units  of  y  down  to  find  the  second  point,  which  helps 
determine  the  line. 


2.  6i 


32/  +  ll  =  0. 

11 


Yk 

J 

rV 

/ 

/ 

io. 

n 

/ 

/ 

/ 

/ 

f 

0 

i 

/ 

' 

3.  x-y  =  0. 

5.  X  +  y  =  4. 

7.  2  X  -  y  =  4. 

9.  Sx-y  =  0. 

11.  x-8y  =  16. 

13.  x  =  8(2-2/). 

15.  X  -  y  -  1  =  0. 

17.  x  +  2/  +  l  =  0. 

19.  12x-3y  =  l. 


4.  x  —  y  —  5  =  0. 

6.  2x  =  6(l-y). 

8.  12x  +  10y  =  5. 
10.  16x-10y  =  4. 
12.  2x  +  y  +  3  =  0. 
14.  2x-6y-l  =  0. 
16.  2x-2y-6  =  0. 
18.  3x-6y-4  =  0. 
20.  7x-8y-9  =  0. 


107.  Solution  of  linear  equations,  and  the  intersection  of  their 
graphs.  The  process  of  solving  a  pair  of  independent  linear 
equations  consists  in  finding  a  pair  of  numbers  (x,  y)  which 
satisfy  them  both.  Though  each  equation  alone  is  satisfied  by- 
countless  pairs  of  values  (x,  y),  we  have  seen  that  there  is  only  one 
pair  that  satisfy  both  equations.    Since  a  pair  of  values  which 


98 


QUADRATICS  AND  BF.YOND 


satisfy  an  equation  are  the  coordinates  of  a  point  on  its  graph,  it 
appears  that  the  pair  of  values  that  satisfy  simultaneously  two 
equations  are  the  coordinates  of  the  point  common  to  the  graphs 
of  the  two  equations,  that  is,  the  coordinates  of  the  points  of 
intersection  of  the  two  lines. 


EXERCISES 

Find  the  solutions  of  the  following  equations  algebraically.   Verify  the 
results  by  plotting  and  noting  the  coor- 
dinates of  the  point  of  intersection. 


1 3^- 

^'  3x- 

-iy 

+  16: 

=  0, 

-y  - 

-7  =  0 

Solution 

3x- 

'4.y 

+  16  = 

0 

3x- 

y 

-    7  = 

0 

Sy 

-23  = 

0  . 

y  = 

¥• 

Substituting 

in  (2), 

3x  = 

7  +  ¥ 

X  = 

(1) 

(2) 


¥• 


To  plot  (1)  and  (2)  we  get  the  equa- 
tions in  the  form  y  =  ax  +  h  and  apply 
the  rule.    Thus 

2/  =  fx  +  4. 

2/  =  3x-7. 


S'^^'iM 

/\l 

^   \ 

i^'f)'! 

y 

"{0.4)'  i 

^^ 

/~ 

z 

/ 

y 

-    i    I 

"7^ 

0           t           X 

.    7 

r 

K'.-i) 

'-t 

. 

JCo.r) 

2    2x  +  3y  =  6, 
7x  +  y  =  2. 


5. 


3x-2y  =  l, 
3x  +  2?/  =  5. 


g    2x  +  y  =  3, 
•  8x-72/  =  l. 

11    x  +  y  =  -i, 
'  4x-3y  =  5. 


14. 


17. 


X  +  y  =  5, 
4x-2y  =  28. 

x-y  =  -4, 
2x  +  6y  =  16. 


6. 

9. 
12. 
15. 
18. 


X  +  y  =  5, 

4 

x-3y  =  -7, 

3x  +  y  =  l. 

4x  +  y  =  ll. 

3x-7y  =  9, 

7. 

X  +  2/  =  -  7, 

X  +  2  y  =  3. 

2x-3y  =  6. 

2x-5y  =  0, 
x-y  =  S. 

10. 

X  +  2  y  =  -  10, 
2x-y  =  0. 

x-y  =  1, 

13. 

x-y  =  l, 

2x-8y  =  3. 

2x-42/  =  -16. 

X  -  2/  =  2, 

16. 

6x-5y  =  5, 

4x-  52/ =  9. 

2  X  +  3  y  =  —  20. 

3x  +  2|/ =  9, 
8x-2/  =  2. 

19. 

2x  +  6y  =  -20,   • 
3x  +  y  =  2. 

GRAPHICAL  REPRESENTATION  99 

108.  Graphs  of  dependent  equations.  We  have  defined  (§  57)  dependent 
equations  as  those  that  are  reducible  to  the  same  form  on  multiplying  or 
dividing  by  a  constant.  Thus  two  dependent  equations  are  reducible  to  the 
same  equation  of  the  form  y  =  ax  +  b.  Hence  dependent  equations  have  as 
their  graphs  the  same  straight  line.  We  see  now  the  geometrical  meaning  of 
the  statement  that  dependent  equations  have  countless  common  solutions. 
Since  their  graphs  have  not  one  but  countless  points  in  common,  being  the 
same  line,  it  is  clear  that  the  coordinates  of  these  countless  points  will 
satisfy  both  equations. 

109.  Incompatible  equations.  By  our  definition  (§  60)  incompatible  equa- 
tions have  no  common  solution.  Since  every  pair  of  distinct  lines  have  a 
common  point  unless  they  are  parallel,  we  can  foresee  the 

Theorem.  Incompatible  equations  have  parallel  lines  as  graphs. 

Let  the  equations 

ax -{- by -{- c  =  0,  (1) 

a'x  +  b'y  +  c'  =  0  (2) 

be  incompatible.    This  is  true  (§  60)  when  and  only  when 

ab'  -  a'b  =  0.  (3) 


(4) 
(5) 

This  may  be  done  if  neither  6  nor  ¥  equals  zero.  If  both  b  and  b'  vanish, 
the  lines  (1)  and  (2)  are  both  parallel  to  the  Y  axis  and  hence  to  each  other, 
which  was  to  be  proved.  But  if  only  one  of  them  vanishes,  say  &  =  0,  then 
by  (3)  a  =  0  (§  5),  in  which  case  (1)  does  not  include  either  variable.  Thus 
we  may  assume  that  neither  b  nor  6'  vanishes  and  that  (4)  and  (5)  may  be 
obtained  from  (1)  and  (2). 

By  (3)  a  _a/ 

b~V 

Our  equations  (4)  and  (5)  become 

a        c 

a        c' 
y  =  --x_-, 

which  represent  parallel  lines,  by  the  Corollary,  p.  96. 

This  theorem  completes  the  discussion  of  the  graphical  representation  of 
the  possible  classes  (§  61)  of  pairs  of  linear  equations. 


Let  us  then  assume  (3). 

Write  (1)  and  (2)  in  the  form 

y  = 

a        c 

—b'-b 

y  = 

100  QUADRATICS  AND  BEYOND 


EXERCISES 


Plot  and  solve : 

1. 

8x  +  2y  =  3, 

4x  +  y  =  8. 

3. 

10x-5y  = 
2  X  -  y  =  3. 

15, 

5. 

x-1y  =  l, 

4x-2Sy  = 

56. 

7. 

x-Sy  =  2, 

6x-lSy  = 

36. 

2x  +  6y  =  1, 
x-^Sy  =  1. 

2x-Sy  =  6, 
8x-  12y  =  24. 

12x-6y=18, 
2x-y=l. 

2x-S  +  y  =  0, 
4x  -7  -\-2y  =0. 


110.  Graph  of  the  quadratic  equation.   Let 

y  =  ax^  -{-  bx  -\-  c,  (1) 

where  as  usual  a,  b,  and  c  represent  integers  and  a  is  positive. 

If  we  let  X  take  on  various  values,  y  will  have  corresponding 

values  and  we  may  plot  the  equation  as  in  §  103.    A  root  of  the 

quadratic  equation 

ax^-}-bx  +  c  =  0  (2) 

is  a  number  which  substituted  for  x  satisfies  the  equation,  that 
is,  gives  the  value  y  =  0  in  (1).  Thus  the  points  on  the  graph 
of  (1)  which  represent  the  roots  of  the  equation  (2)  are  the 
points  for  which  y  =  0,  that  is,  where  the  curve  crosses  the 
X  axis.  The  numerical  value  of  the  roots  is  the  measure  of 
the  distance  along  the  X  axis  from  the  origin  to  the  points  where 
the  curve  cuts  the  axis.  Since  this  distance  is  always  a  real 
distance,  only  real  roots  are  represented  in  this  manner. 

Theorem.  If  the  graph  of  (1)  has  no  point  in  common  with 
the  X  axiSf  the  equation  (2)  has  imaginary  roots,  and  conversely. 

Every  equation  of  form  (2)  has  two  roots  either  real  or  imagi- 
nary (§  89).  If  the  graph  of  (1)  has  no  point  in  common  with 
the  X  axis,  there  is  no  real  value  of  x  for  which  y  =  0,  i.e.  no 
real  root  of  (2).    The  roots  must  then  be  imaginary. 

Conversely,  if  (2)  has  only  imaginary  roots,  there  is  no  real 
value  of  X  which  satisfies  it,  i.e.  which  makes  y  =  0  in  (1). 
Thus  the  curve  has  no  point  in  common  with  the  X  axis. 


GRAPHICAL  REPRESENTATION  1,01 

This  suggests  the  following  universal 

Principle.  Non-intersection  of  graphs  corresponds  to  imagi- 
nary or  infinite-valued  solutions  of  equations. 

111.  Form  of  the  graph  of  a  quadratic  equation.   Consider  the 

equation  ^    o      rr         ^ 

^  y  ^2x^+1  x-\-2.         ^  (1) 

By  substituting  for  x  a  very  large  positive  or  negative  number, 
say  X  =±  100,  y  is  large  positively.  Thus  for  values  of  x  far  to 
the  right  or  left  the  curve  lies  far  above  the  X  axis.  If  we 
assign  to  t/  a  certain  value,  say  3/  =  2,  we  can  find  the  correspond- 
ing values  of  x  by  solving  a  quadratic  equation.    Thus  in  (1)  let 

or       .  2x^  +  1  x  =  0. 

The  roots  are  x^=—  3^,    x^  =  0. 

Hence  the  points  (—  3^,  2)  and  (0,  2)  are  on  the  curve  (§  101). 
That  is,  if  we  go  up  two  units  on  the  Y  axis,  the  curve  is  to  be 
found  three  and  one  half  units  to  the  left  and  also  again  on  the 
Y  axis.  If  in  (1)  we  let  y  =  —  4,  the  corresponding  values  of  x 
are  very  nearly  equal  to  each  other  (—  1^  and  —  2),  which  means 
that  the  curve  meets  a  line  parallel  to  the  X  axis  and  four  units 
below  it  at  points  very  near  together.  The  question  now  arises, 
Where  is  the  bottom  of  the  loop  of  the  curve  ?  This  lowest  point 
of  the  loop  has  as  its  value  of  y  that  number  to  which  correspond 
equal  values  of  x.  Hence  we  must  determine  for  what  value  of 
y  the  equation  (1),  that  is,  the  equation 

2x^-{-lx  +  {2-y)=0, 

has  equal  roots.    Comparing  with  the  equation  ax"^  +  Jx  +  c  =  0 


or 


Z  - 

=  a,  7  = 

--b,2- 

-y=-c. 

Thus  the  condition  h 

2-4ac 

=  0  becomes 

49-4 

•2(2- 

-2/)=0, 

y  = 

49- 
8 

16_ 

33 

"  8 

4i. 

102 


QUADRA'EICS  AND  BEYOND 


I  as  the  corre-' 


Substituting  this  value  of  y  in  (1),  we  get 
spending  value  of  x. 

This  gives  a  single 
value  of  y  for  which 
the  values  of  x  are 
equal;  hence  the  graph 
of  (1)  is  a  single  fes- 
toon as  in  the  figure. 

If  we  take  the  gen- 
eral equation 

ax^  -{-  bx  -{-  c  =  y, 

we  find  precisely  similarly  that  the  bottom  of  the  loop  is  at  a 
point  whose  ordinate  is 

P  —  Aac  A 

y  =  - 


y 

X 

0 

2 

-4 

-.3+ or -3.2  + 
0  or  -  ^ 
-  li  or  -  2 
-If 

4a 


4a 


Thus  we  see  again  that  if  the  discriminant  is  negative  the  graph 
is  entirely  above  the  X  axis  and  both  roots  are  imaginary  (§§  98, 
110),  since  the  ordinate  of  the  lowest  point  of  the  loop  is  positive. 
If  the  discriminant  is  positive,  the  graph  cuts  the  X  axis  and  both 
roots  are  real. 

The  results  of  this  section  enable  us  to  determine  a  value  of 
y  from  the  coefficients  which  determine  the  lowest  point  of  the 
loop  of  the  curve  precisely,  and  hence  to  show  beyond  question 
from  the  graph  whether  the  equation  ^as  real  or  imaginary  roots. 


EXERCISES 

Plot  the  following  equations  and  determine  by  measurement  the  roots  in 
case  they  are  real.    Find  in  each  case  the  lowest  point  on  the  loop. 


1.  x2  +  X  +  1  =  y. 

4.  x2  +  7  aj  +  6  =  y. 

7.  x2-6x  +  l  =  y. 

10.  x2  +  2x-l  =  2/. 

13.  2x2-x-3  =  y. 


3.  x2-6x  +  10  =  ?/. 
6.  3x2-7x-6  =  y. 
9.  2x2-9x  +  7  =  y. 


2.  x2  -  4  X  +  7  =  y. 

5.  x2-6x  +  9  =  y. 

8.  x2-6x  +  5  =  y. 

11.  x2  -4x  +  4  =  y. 

14.  3  x2  +  8  X  +  5  =  y. 

16.  What  is  the  characteristic  feature  of  the  plot  of  an  equation  whose 

roots  are  equal  ? 


12.  3: 


4x-3  =  y. 


15.  4x2+12x+9  =  y. 


* 


GRAPHICAL  REPRESENTATION  103 

112.  The  special  quadratic  adc^  +  6ic  =  O.   When  in  the  quad- 
ratic equation 

ax^  -\-hx^  G  =  0,  (1) 

c  =  0,  we  can  always  factor  the  equation  into 

ax^  -\-  bx  =  X  (ax  -j-  b)  =  0, 


(—:)=»■ 


or 

Thus  the  roots  are  £Ci  =  0,  X2= 

Conversely,  if  a;  '=  0  is  a  root,  then  (§  95)  ic  —  0,  or  x,  is  a  factor 
and  the  equation  can  have  no  constant  term. 
This  affords  the 

Theorem.    A  quadratic  equation  has  a  root  eqtcal  to  zero 
when  and  only  when  the  constant  term  vanishes. 

We  show  in  a  similar  manner  that  both  roots  of  the  equation 
(1)  are  zero  when  and  only  when  i  =  c  =  0. 

EXERCISES 

1.  Prove  the  theorem  just  given  by  considering  the  expressions  for  the 
roots  in  terms  of  the  coefficients  (§  89). 

2.  For  what  real  values  of  k  do  the  following  equations  have  one  root 
equal  to  zero  ? 

(a)  x2  +  6x  -  A:  +  1  =  0.  (b)  2x2  -  3x  +  A;2  -  1  =  0. 

(c)  x2  +  6a;  ^.  ^2  ^.  1  ^  0.  (d)  2x2  -  4x  +  fc2  _  3^  ^  0. 

(e)  2x2  +  2fcx-2A;2-4A;-2=0.     (f)  6x2  -  4x  +  2 A;2  +  fc  +  7  =  0. 

3.  What  is  the  characteristic  feature  of  the  plot  of  an  equation  which  has 
one  root  equal  to  zero  ? 

4.  For  what  real  value  of  A:  will  both  roots  of  the  following  equations  vanish  ? 

(a)  -  +  3x  -  1  =  0.  (b)  x2  +  (fc2  +  3)x  +  fc  -  3  =  0. 

(c)  x2  +  (fc2  +  l)x  +  l  =  0.  (d)  x2+(fc-3)x  +  2fc2-5fc-3  =  0. 

(e)  x2  +  (A:  +  l)x  +  /c2 -1  =  0.         (f)  (fc-3)x2  +  (fc2_9)x  +  A;2-4A;  +  3=a 


104  QUADRATICS  AND  BEYOND 

113.  The  special  quadratic  ax^  +  c  =  O.    This  equation  may 
be  written  in  the  form  x^  -\-  -  =  0  and  factored  *  immediately  into 


(^-*-^R)(^-^R)=o> 


which  shows  that  the  roots  are  equal  numerically  but  have  oppo« 
site  signs.    The  roots  are 


Xi   =\/-^'      ^2 


xR 


Since  in  the  equation  ax^  -\-  c  =  y  the  variable  x  occurs  only  in 
the  term  a:^,  we  get  the  same  value  of  y  for  positive  and  negative 
values  of  x.  Hence  the  loop  which  forms  the  graph  of  the  equa- 
tion is  symmetrical  with  respect  to  the  Y  axis. 

114.  Degeneration  of  the  quadratic  equation.   The  equation 

ax^  -\-hx  -\-  c  =■  0 
has  the  roots 


_  J  4_  V52  -  4  ac 

Xi  — 

2a 

Xo   = 

We  wish  to  find  the  effect  on  the  roots  x^  and  x^  when  a 
becomes  very  small.    If  we  let  a  approach  0,  then  x^  approaches 

an  expression  of  the  form  -?  which  must  always  be  avoided. 

Rationalize  the  numerators  and  we  get 

4  ac  2  c 


2a{-  b  -  ■\/b^  -  4  ac)       -b-\^b^-4:  ac 
4  ac  2  c 


2a{-b  +  -^b^-4:ac)      -  b -{-  ^b^  -4.ac 
As  a  approaches  0,  evidently  h^  —  ^ac  approaches  J^,  Xi  ap- 
proaches  —  j>  and  x^j  since  its  denominator  becomes  very  small, 

•  When  —  Is  positive  this  involves  real  factors.    If  —  is  negative  the  factors  are 
a  a 

imaginary  (§  152). 


GRAPHICAL  REPRESENTATION 


105 


increases  without  limit,  that  is,  approaches  infinity.  Thus  the 
quadratic  equation  approaches  a  linear  equation  when  a  approaches 
0,  and  one  of  its  roots  disappears  since  it  has  increased  in  value 
beyond  any  finite  limit.  The  loop-shaped  graph  of  the  quadratic 
equation  must  then  approach  a  straight  line  as  a  limit  when  a 
approaches  0.  This  is  made  clear  from  the  following  figure,  where 
a  has  the  successive  values  1,  \,  y^^,  ^V,  0. 

In  the  figure  the  curves  represent  the  following  equations : 


x-'-l-2  =  y.. 

(I) 

x"       X 
5-2-2  =  ^- 

(H) 

10-2-2  =  2'- 

(III) 

X^       X        ^ 

50-2-2  =  ^- 

(IV) 

-f-2- 

(V) 

%%-A  "■ 

Ml  II 171 

^^^J 

7          1 

tsX^ 

t     J~ 

>^  \^v  ^ 

t      t     h- 

^§^^A 

l        f     -  ^ 

^SSv^ 

t          ^      y^^^ 

^^> 

^  ^'^,^ 

< 

N^^ 

^^^>^ 

^^^-L*- 

^-.K 

In  a  similar  manner  we  can  show  that  when  in  the  equation 
Jx  +  c  =  0,  &  approaches  0  as  a  limit,  the  root  of  the  linear  equa- 
tion becomes  infinite. 


EXERCISES 

1.  "What  real  values  must  k  approach  as  a  limit  in  order  that  one  root  of 
each  of  the  following  equations  may  become  infinite  ? 


(a)  Arx^  +  Gx  +  l  =  0. 

(c)  (A;x  - 1)2  -  (X  +  2)2  =  (A:  +  a:)2. 

(e)   V2A:x-H-\/6F=^=V^TT. 

X 


(b)  (^2  +  1)3524.  a;  + 1  =  0. 
(d)  A;2  +  4fc2a;2_(aj_i)2  4.2 


(g) 


1      fc  +  1       X 
1      X  + 1      A;2 


0. 


(f )  Vx-fc  +  Vx  +  A;  =  yfkx  +  1. 
1~\     fc2 


/fcx-1 
'a:x  +  1 


(i) 


(^ 


1)^  ^  (fc 


(A;  +  1)      k 


(X  +  1)2 
(j)  (A:2  -  l)x2  +  (A:  -  l)x  +  A:2  +  4A;  -  5 


106  QUADRATICS  AND  BEYOND 

2.  What  real  values  must  k  and  m  approach  as  a  limit  in  order  that  both 
roots  of  the  following  may  become  infinite  ? 

(a)  A:x2  +  mx  +  1  =  0. 

(b)  (2fc-m)x2  +  A:x-2  =  0. 

(c)  2 tec2  +  (3m  -  1  +  h)x  =  8x2  -  1. 

(d)  ijc  -  l)x2  +  (A;  +  m  +  l)x  +  3  =  0. 

(e)  x2  -  X  -  2(fc  +  m)a;  =  (fc  +  m)  (x2  -  1). 

(f)  (fc  +  m)x2  +  2 (fc  +  m)  +  1  =  x2  -  2 x. 

(g)  (fc  +  m  +  l)x2  +  (2  A:  -  m  -  l)x  +  1  =  0. 
(h)  (2A;  +  m  +  2)x2  +  (4A:  +  2m  +  3)x  +  3  =  0. 

115.  Sum  and  difference  of  roots.    Let  x^  and  x^  be  the  roots  of 

x''  +  bx  +  G  =  0.  (1) 

Then  (§  95)  x  —  x^  and  x  —  x^  are  factors,  and  their  product 
^'^  —  (^1  +  ^2)^  +  ^1^2  is  exactly  the  left-hand  member  of  (1). 
Consequently  the  equation 

x^  -\-  bx  -\-  c  =  x^  —  (xi -{-  Xz)  X  +  X1X2 
is  true  for  all  values  of  x.    Hence  by  §  96 

-  (^1  4-  X,)  =  b,  (2) 

X1X2  =  c.  (3) 

We  may  state  these  facts  in  the 

Theorem.    The  coefficient  of  x  in  the  equation  s(^-^hx-\-c=0* 
is  equal  to  the  sum  of  its  roots  with  their  signs  changed. 
The  constant  term  is  equal  to  the  product  of  the  roots. 

EXERCISES 

1.  Prove  the  statement  just  made  from  the  expression  for  the  roots  in 
terms  of  the  coefficients  (§  89). 

2.  Form  the  equations  whose  roots  are  the  following : 

(a)  6,  1.  (b)  i,  I  (c)  I,  3.  (d)  -  i  -  6. 

(e)  h  h  (f)   -h+h  (g)  2  +  V3,  2  - V3.         (h)   - V3,  V3. 

*  We  should  for  the  present  exclude  the  case  where  6^  -  4c<0,  since  the  roots  ar,  and 
a?j  are  then  imaginary  and  we  have  not  as  yet  dettned  what  we  mean  by  the  sum  or 
the  product  of  imaginary  numbers.  We  shall  see  later  that  the  theorem  is  also  true  in 
this  case. 


GRAPHICAL  REPRESENTATION  107 

3.  If  4  is  one  root  of  ic^  —  3  x  +  c  =  0,  what  value  must  c  have  ? 
Solution  :  Let  Xi  be  the  remaining  root. 

Then  by  (1)  _  (xj  +  4)  =  -  3, 

or  xi  =  —  1. 

By  (2)  c  =  Xi.4=:(-l)4  =  -4. 

4.  Find  the  value  of  the  literal  coefficients  in  the  following  equations. 

(a)  x^-\-bx-9  =  0.     One  root  is  3. 

(b)  0:2  +  4  X  +  c  =  0.     One  root  is  2. 

(c)  ax2  + 3x-4  =  0.     One  root  is  2. 

(d)  ax2  +  3x  +  4  =  0.     One  root  is  7. 

(e)  ax2  +  2  X  +  6  =  0.     One  root  is  6. 

(f )  x2  +  6x  +  4  =  0.     One  root  is  -  1. 

(g)  x2  -  6x  -  6  =  0.     One  root  is  -  3. 
(h)  x2  +  6x  +  6  =  0.     One  root  is  -  6. 

(i)  2  x2  —  6  X  —  c  =  0.     One  root  is  —  4. 

(j)  x2  —  6  X  +  c  =  0.     One  root  is  double  the  other. 

(k)  x2  +  c  =  0.     The  difference  between  the  roots  is  8. 

(1)  x2  —  5  X  +  c  =  0.     One  root  exceeds  the  other  by  3. 

(m)  x2  —  7  X  +  c  =  0.     The  difference  between  the  roots  is  6. 

(n)  x2  —  C  X  +  c  =  0.     The  difference  between  the  roots  is  4. 

(o)  x2  —  3  X  +  c  =  0.     The  difference  between  the  roots  is  2. 

(p)  x2  —  2  X  +  c  =  0.     The  difference  between  the  roots  is  8. 

116.  Variation  in  sign  of  a  quadratic.   It  is  often  necessary  to  know  the 

sign  of  the  expression 

ax2  +  6x  +  c 

for  certain  real  values  of  x,  and  to  determine  the  limits  between  which  x  may 
vary  while  the  expression  preserves  the  same  sign.  We  assume  as  usual  that 
a  is  positive. 

Theorem  L*  If  the  discriminant  of  ax^  +  hx  +  cis  positive,  the  quadratic 
is  negative  for  all  values  of  x  between  the  values  of  the  roots  of  the  equation.  For 
other  values  of  x  {excepting  the  roots)  the  quadratic  is  positive. 

*  If  a  were  negative,  Theorem  I  would  read  as  follows  :  If  the  discriminant  is  posi- 
tive, the  quadratic  is  positive  for  all  values  of  x  between  the  values  of  the  roots  of  the 
equation.    For  other  values  of  x  {excepting  the  roots)  the  quadratic  is  negative. 

When  a  is  negative  Theorems  II  and  III  may  be  modified  in  an  analogous  manner. 


108 


QUADKATICS  AND  BEYOND 


\ 


In  §  98  we  found  that  when  the  discriminant  of  a  quadratic  equation  is 
positive  the  equation  has  two  real  roots.  If  two  roots  are  real,  the  loop  of 
the  graph  of  the  equation  ax^  +  bx  +  cz=y  cuts  the  X  axis  in  two  points 

(§  110)  as  in  the  figure.  The  roots  are 
represented  by  A  and  B,  and  any  real 
value  of  X  between  the  roots  is  repre- 
sented by  a  point  P  in  the  line  J.JB. 
Since  the  curve  is  below  the  X  axis  at 
any  such  point,  the  value  of  y,  i.e.  of 
the  expression  ax^  +  &x  +  c  for  values 
of  X  between  the  roots,  is  negative. 

The  value  of  the  expression  for  any 
value  of  X  greater  or  less  than  both 
roots  is  seen  to  be  positive,  since  for 
such  points,  for  example  Q  and  R,  the  graph  is  above  the  X  axis. 

Theorem  II.  If  the  discriminant  of  ax^  -\-  hz  +  c  is  negative,  the  expression 
positive  for  all  real  values  of  x. 

When  the  discriminant  is  negative  the  entire  graph  of  ax^  +  &x  +  c  =  y  is 
above  the  X  axis  (§  111),  and  consequently  for  any  real  value  of  x  the  corre- 
sponding value  of  y,  i.e.  the  value  of  ax^  +  &x  +  c,  is  positive. 

Theorem  III.  If  the  discriminant  of  ax^  +  bx  +  c  is  zero,  the  value  of  the 
expression  is  positive  for  all  values  of  x  except  the  roots  of  the  equation 
ax^  -\-  bx  +  c  =  0. 

Hint.  See  example  16,  p.  102. 

We  may  restate  these  three  theorems  and  prove  them  algebraically  as 
follows : 

Theorem  IV.  If  the  discriminant  of  the  quadratic  ax^  +  bx-\-  cis  positive, 
the  values  of  the  quadratic  and  a  differ  in  sign  for  all  values  ofx  lying  between 
the  roots,  and  agree  for  other  values. 

If  the  discriminant  is  zero  or  negative,  the  value  of  the  quadraiic  always 
agrees  with  a  in  sign. 

Case  I.  Since  the  discriminant  is  positive,  the  equation  ax^+bx+c=(i 
has  two  unequal  real  roots,  as  Xi  and  x^,  of  which  we  will  assume  Xi  is  the 
greater,  and  we  may  write  the  quadratic  in  the  form 

ax2  +  6x  +  c  =  a  (x  —  Xi)  (x  —  X2). 

Now  for  any  value  of  x  between  Xi  and  X2  the  factor  x  —  Xi  is  negative, 
while  X  —  Xa  is  positive,  which  shows  that  the  quadratic  is  opposite  in  sign 
to  a  for  such  values  of  x.  For  other  values  of  x  both  these  factors  are  either 
positive  or  negative,  and  for  such  values  the  quadratic  is  of  the  same  sign 
as  a. 


GKArillCAL  KEPRESENTATION  109 

Case  II.    Since  the  discriminant  6^  _  4  ^c  is  negative  and  the  roots  are 

*  *!,    *  &  ±  V&2  -  4  ac  .^    ^^  ,    -: . 

of  the  form  —  — — — - ,  we  may  write  the  quadratic 

2a 

„  ,  ,      ,  /         &       V62  -  4  ac  V     ,    6    .    V62  -  4  ac\ 

V        2a  2a       /\       2a  2a       / 

Now  for  any  value  of  x  the  expression  lx-\ )    is  positive,  and  since 

-     \        2  a/ 
62  — 4ac  is  negative,  4ac  — 6^  is  positive;    and  we  observe  that  tlie  last 
member  of  the  equation  has  the  same  sign  as  a. 

Case  III.    Since  the  discriminant  is  zero,  the  roots  are  equal  and  the 
expression  has  the  form 

ax2  4-  &x  +  c  =  a  (x  —  Xi)2, 

which  has  evidently  the  same  sign  as  a,  for  any  value  of  x. 

EXERCISES 


1.  Between  what  values  of  x  is  the  expression  Vx^  —  5  x  +  4  imaginary  ? 

Solution :  The  roots  of  x^  —  5  x  +  4  =  0  are  4  and  1. 

The  discriminant  A  =  62  _  4  ^c  =  25  —  16  =  9  is  positive. 
Thus  by  Theorem  I  or  IV,  if  1  <  x  <  4  f  the  expression  under  the  radical 
sign  is  negative  and  the  whole  expression  is  imaginary. 

2.  For  what  values  of  k  are  the  roots  of 

(A:+ 3)x2  +  A:x  +  l  =  0  (1) 

(a)  real  and  unequal  ?  (b)  imaginary  ? 

Solution :  a  =  k  +  S,  b  =  k,  c  =  1. 

A  =  b^-4ac  =  k^-  4{k  +  S)  =  k'^-ik-  12. 

(a)  If  A  >  0,  the  roots  of  (1)  are  real  and  unequal. 
The  roots  of  A;^  -  4  A:  -  12  are  A:  =  -  2  and  6 
Then,  by  Theorem  II,  if 

A;  <  -  2  or  fc  >  6,  A  >  0. 

(b)  By  Theorem  I,  if         -  2  <  A;  <  6,  A  <  0, 
and  the  roots  of  (1)  are  imaginary. 

*  See  5 162.       t  Road  "  1  is  le6S  than  x  wliich  is  lew  than  4  "  or  "  a;  is  between  1  and  4." 


110  QUADRATICS  AND  BEYOND 

3.  Determine  for  what  values  of  x  the  following  expressions  are  negative, 
(a)  aj2  +  2x  -  1/  (b)  x^-Sx-h  4. 

(c)  x2  -  11  oj  +  10.  (d)  x^-15x-{-  60. 

(e)-x2-2x  +  l.  {i)  -  x^ -\- 7 X  +  SO. 

4.  Determine  for  what  values  of  k  the  roots  of  the  following  equations 
are  (a)  real  and  unequal,  (b)  imaginary. 

(a)  3  A;x2  -  4  X  -  2  =  0.  (b)  aj2  +  4  fcx  +  A:2  +  1  =  0. 

(c)  x^-{-9kx  +  6k  +  l  =  0.  (d)  x2  +  {3k  +  l)x  +  1  =  0. 

(e)  (A;2  +  3)x2  +  A;x  -  4  =  0.  (f)  2 x2  -  4x  -  2  A;  +  3  =  0. 

(g)  fcx2  +  (4A;  +  l)x  +  4  A:  -  3  =  0.       (h)  {k  -  l)x2  +  6kx -{- 6k  +  i  =  0. 
(i)  {k  -  l)x2  4-  {2k+l)x  +  A:  +  3  =  0. 


CHAPTEE  X 

SIMULTANEOUS  QUADRATIC  EQUATIONS  IN  TWO 
VARIABLES 

117.  Solution  of  simultaneous  quadratics.  A  single  equation 
in  two  variables,  as  x^  -\-  y^  =  5,  is  satisfied  by  many  pairs  of 
values,  as  (1,  2),  (Vl>  V|)>  (2,  1),  and  so  on,  though  there  are 
at  the  same  time  numberless  pairs  of  values  that  do  not  satisfy 
it,  as  (0, 1),  (1, 1),  (2,  3).  Thus  the  condition  that  (x,  y)  satisfy  a 
single  quadratic  equation  imposes  a  considerable  restriction  on 
the  values  that  x  and  y  may  assume.  If  we  further  restrict  the 
value  of  the  pair  of  numbers  (cc,  y)  so  that  they  also  satisfy  a 
second  equation,  the  number  of  solutions  is  still  further  limited. 
The  problem  of  solving  two  simultaneous  equations  consists  in 
finding  the  pairs  of  numbers  that  satisfy  them  both. 

118.  Solution  by  substitution.  In  this  method  of  solution  the 
restriction  imposed  on  (x,  y)  by  one  equation  is  imposed  on  the 
variables  in  the  other  equation  by  substitution. 

Example.    Solve  2x^-\-y'^-l,  (1) 

x-y  =  \.  (2) 

Solution  :  Equation  (2)  states  that  x  =  1  +  y.  Thus  our  desired  solution  is 
such  a  pair  of  numbers  that  (1)  is  satisfied  and  at  the  same  time  x  is  equal 
to  y  +  1. 

If  we  substitute  in  (1)  1  +  y  for  x,  we  are  imposing  on  its  solution  the 
restriction  implied  by  (2). 

Thus  2(l  +  2/)2  +  2/2=,i^ 

or  \  3  2/2  4-  4  y  +  1  =  0. 

The  roots  are  y  =  —  1,   y  =  —  i. 

Corresponding  to  ?/  =  —  1  we  get  from  (2)  x  =  0. 
Corresponding  to  y  =—  \  we  get  from  (2)  x  =  |. 
Thus  the  solutions  are  (0,  —  1)  and  (|,  —  \). 

Ill 


112  QUADRATICS  AND  BEYOND 

EXERCISES 

Solve  the  following : 


xy  =  4. 

2. 

x-y  =  6, 
xy  =  36. 

■  x2  +  2/2  =  bxy. 

xy  —  x  =  0. 

5. 

x-hy  =  a, 

x2  +  2/2  =,  5. 

x2  4-2/2  =  50, 
•  9x  +  7  2/  =  70. 

'•  x-Sy  =  0. 

8. 

2x-3y  =  4, 
x2  -  y2  =  0. 

9    xy  =  12, 

2x  +  3?/  =  18. 

10    x:y  =  9:4, 
^"-  a;:12  =  12:y. 

11. 

X2:y2=:a2:62^ 
a  —  X  =  b  —  y. 

12    5x2  +  2/  =  3xy 
2x-2/  =  0. 

*"•  2x-32/  =  0. 

14    (a^  +  y)C 

z-2y)  =  T, 
3. 

-g    3x2-4y  =  6x- 
3x+42/  =  10. 

2y% 

16   x2  +  ,= 
x:y  =  2 

2/2  +  X  -  18, 
:3. 

^^    ax-by  =  cy,  ^g    x2  +  2 X2/  +  2/2  =  7  (x  -  y), 

a2x2  —  62y2  _  acx2/  +  to2.  '  2  X  —  2/  =  5. 

^g    ax2  +  (a-6)x2/-&2/=^=c2,  ^q    2x2-5x2/+2/2+10x+12y=100, 

■   (x  +  2/) :  (X  -  2/)  =  a :  6. 

2j    7(x  + 5)2 -9(2/ +  4)2  =  118, 
*   X  -  2/  =  1. 


x2  +  2/2  =  130, 
23.  X  +  y      ^ 


x-y 

2X-2/  +  1      8 

25. 

x-22/  +  l~3' 

x2  -  3  X2/  +  2/2  =  5. 

27. 

X2/  +  72  =  6(2x  +  2/), 
X      2 

V     8 

4X+2/-1      4X+2/-12 

29. 

2X+2/-1      2X+2/-12 

3x  +  y  =  13. 

31. 

10            9' 
x  +  2  +  2/-l  =  '' 
2         4 

=  2f, 


2x-32/  =  l. 

22. 

x2  +  2/2  =  a2, 
X      m 

2/      n 

24. 

x2  +  2/  +  1      3 
2/2  +  X  +  1      2 

x  —  y  =  l. 

26. 

1  +  X  +  X2^3 

1  +  2/  +  2/^ 
a;  +  2/  =  6.  • 

28. 

2/  +  6       X 

X  —  2/  =  4. 

30. 

«(x-2/)-52/  =  6 
x-2/ 

32. 

?  +  ?  =  3, 

x      2/ 

^^Tri  =  ;-  5(2/-l)  =  2(x  +  l). 


SIMULTANEOUS  QUADRATIC  EQUATIONS  113 

119.  Number  of  solutions.  We  have  proved  (p.  42)  that  two 
linear  equations  have  in  general  one  and  only  one  solution. 

Theorem.  A  quadratic  equation  and  a  linear  equation  ham 
in  general  two  and  only  two  sohUions. 

If  the  linear  equation  is  solved  for  one  variable,  say  x,  and  this 
is  substituted  in  the  quadratic  equation,  we  get  a  quadratic  equa- 
tion to  determine  all  possible  values  of  the  other  variable  (i.e.  y), 
which  must  in  general  be  two  in  number  (§  98).  To  each  one  of 
these  values  of  y  will  correspond  one  and  only  one  value  of  x, 
thus  affording  two  solutions  of  the  pair  of  equations. 

EXERCISES 

1.  When  may,  as  a  special  case,  a  quadratic  and  a  linear  equation 
have  only  one  solution? 

2.  When  may  a  quadratic  and  a  linear  equation  have  imaginary 
solutions  ? 

3.  rind  the  real  values  of  k  for  which  the  following  equations  have 
(1)  only  one  solution,  (2)  imaginary  solutions. 

x2  +  y2  =  16,  (1) 

^^^  x-y  =  k,  (2) 

Solution :  x  =  y  +  k. 

Substitute  in  (1),         {y  +  fc)2  -{- y"^  =  16. 
or  2  y2  4-  2  A:y  +  A;2  -  16  =  0. 

As  in  §  98,  a  =  2;  b  =  2k;  c  =  k^ -16. 

Hence  A  =  b'^  -  Aac  =  ^k^  -  Sk^  +  12S  =  -  Ak^  +  128.    • 

(1)  A  =  0  when  4k^  =  128, 

or  k  =  ±  4  -^2.    There  is  then  only  one  solution. 

(2)  A  <  0     when     k^  >  32.    The  solution  is  then  imaginary. 

^^  2x-y  =  k. 

x^-y^  =  9, 
^^^  x-2y  =  k. 

2x2  +  3y2  =  6, 
^J^  x-ky  =  \. 


x  +  ky  =  b, 

^""^  (C2  +  2/2  =  5. 

^    y-x  =  k, 

^  ^  x^y  =  k. 

x2  +  2/2  =  25, 
(')  4x-3y  =  k. 

114  QUADRATICS  AND  BEYOND 


1 


.  120.  Solution  when  neither  equation  is  linear.  In  the  exam- 
ples previously  given  one  equation  has  been  linear  and  the  other 
quadratic  in  one  or  both  variables.  Often  when  neither  of  the 
original  equations  is  linear  a  pair  of  equivalent  (p.  41)  equations 
one  or  both  of  which  are  linear  may  be  found.  These  latter  equa- 
tions may  be  solved  by  substitution. 


EXERCISES 

Solve  the  following  equations. 

When  neither  equation  is  linear,  we  can  often  obtain  by  addition  an 
equation  from  which  by  the  extraction  of  the  square  root  a  linear  equation 
may  be  found. 

1.  x2  +  2/2  =  17,  (1) 

xy  =  4.  (2) 

Solution:  x^  +  y^  =  17  (3) 

Multiply  (2)  by  2,         2xy  =    8  (4) 

Add  x2  +  2  icy  +  2/2  ^  25 

Extract  the  square  root,  a;  +  y  =  ±  5.  (5) 

Subtract  (4)  from  (3),  x2  -  2  xy  +  2/2  =  9. 

Extract  the  square  root,  x  —  y  =±S.  (6) 

Solve  (5)  and  (6)  as  simultaneous  equations, 

X  +  y  =  ±6, 

x~y  =  ±S. 

x  =  +i,  +1,  -1,  -4. 
2/  =  +l,   +4,   -4,   -1. 
Thus  the  solutions  are  four  in  number,  (4,  1),  (1,  4),  (—1,  —  4),  (—  4,  —1). 

The  following  exercise  affords  another  case  where  a  linear  equation  may 
be  found  by  addition  and  extraction  of  the  square  root. 

2.  x^-hxy  =  6,  (1) 
xy  +  y^  =  10.                                               (2) 

Solution :  Add  (1)  and  (2), 

x^-^2xy  +  y^  =  16. 
Extract  the  square  root,  «  +  2/  =  ±  4. 

Substitute  in  (1),       x2  +  x  ( ±  4  -  x)  =  6, 
x2±4x-x2=:6, 

X  =  ±  f  =  ±  1^. 

Substitute  in  (4),  2/  =  2^,  —  2|. 

Thus  our  solutions  are  (-  |,  -  2|),  (f,  +  2J). 


SIMULTANEOUS  QUADRATIC  EQUATIONS  115 

When  neither  of  the  original  equations  is  quadratic,  we  can  often  find  by 
division  an  equivalent  pair  of  equations  one  of  which  is  linear  and  the  other 
quadratic,  as  in  the  following  exercise. 

3.  x3  +  2/8=12,  (1) 

X  +  2/  =  2.  (2) 

Solution:  Divide  (1)  by  (2), 

x2  -  ajy  +  y2  ==:  6.  (3) 

Square  (2),  x^  +  2  xy  +  y^  =  4 

Subtract,  —  3  xy  =  2 

xy=-l  (4) 

Solve  (4)  with  (2)  by  substitution. 

When  the  sum  of  the  exponents  of  the  variables  is  the  same  in  every 
term,  the  equation  is  called  homogeneous. 

Thus,  z'^  +  xy^O,        2x^y -3zy^  - 'iz^  -  3y^  =  0. 

When  one  equation  is  homogeneous  and  the  other  either  linear  or  quadratic 
we  may  solve  them  as  follows  : 

4.  6x2-7xy  +  2y2  =  o,  (1) 

x2  -  y  =  4.  (2) 

Solution :  Divide  (1)  by  y^, 

Let-  =  2,«  622_7z  +  2  =  0. 

y 

Solve  for  z,  z  =  |  or  |. 

Thus  ?=:ior?  =  ?. 

y     2        y      3 

Solve  (2)  with  2  x  =  y  and  3  x  =  2  y. 

When  both  equations  are  homogeneous  except  for  a  constant  term  we  may 
solve  as  follows : 

5.  x2-x2/  +  2y2  =  4,  (1) 
2x2-3x!/-2?/2  =  6.  (2) 

Solution :  Eliminate  the  constant  term  by  multiplying  (1)  by  3  and  (2)  by  2, 
3x2-3x2/+    6?/2  =  i2,  (3) 

4  x2  -  6  xy  -    4  y2  =  12  (4) 

Subtract  (3)  from  (4),       x2  -  3  xy  -  10  y2  =   o 

*  "We  observe  that  y  ^0.  For  if  y  =  0  were  a  value  that  satisfies  equation  (1),  x  =  0 
would  correspond.  But  (0, 0)  does  not  satisfy  (2);  thus  y  =  Ois  not  a  value  that  can  occur 
in  the  solutions  of  the  equations. 


116  QUADRATICS  AND  BEYOND 


X 

Divide  by  y^  and  let  -  =  2,  where  y  ^d, 

y 

22  _  32  _  10  =  0.  (S)' 

Factor,  (z  -  6)  (2;  +  2)  =  0. 

The  roots  are  _  =  6,  -  =  -  2.  (6) 

y         y 
Solve  (6)  with  (1). 

"When  one  equation  is  quadratic  in  a  binomial  expression  we  may  solve  as 
follows : 


6.                                       x-y 

-Vx-2/  =  2,                                             (1) 

x^-ys  =  2044.                                       (2) 

Solution:  Let 

Vx  -  ?/  =  z. 

Then  (1)  becomes 

z2  -  z  =  2. 

Solving  for  z. 

z  =  2  or  -  1. 

Thus 
or 

Solve  (3)  with  (2)  as  in  exercise 

x-2/  =  n                                          (3) 
x-y  =  lj                                        ^  ' 

3. 

7.  x^  +  y^  =  xy  =  X  -\-  y. 

8.  X3  +  2/3  =7x2/  =  28(x  +  2/). 

2/3  +  x^y  =  4. 

x2^  =  a, 
^"-   X2/2  =  b. 

-      x{y-l)  =  10, 
'  y{x-l)  =  12. 

12    a^2  +  2/2-a, 
'  xy  =  b. 

^^    x^y  +  xy^  =  a, 

x^y  —  xy^  -  b. 

..    x  +  X2/=:35, 
•  2/  +  X2/  =  32. 

-g    x{x^-{-y^)  =  7, 
y{x^-\-y^)  =  l. 

2x2-32/2  =  6, 
•  3x2-2  2/2  =  19. 

3x2-22/2  =  16. 

-J.    5x2  +  2^2  =  22, 
3x2-52/2  =  7. 

IQ    ic2  +  ccy  +  2/2  =  2, 
•  x2  -  X2/  +  2/2  =  6. 

20    x  +  X2/  +  2/  =  5, 

■  a;2  +  X2/  +  2/2  =  7. 

Hint.  Eliminate  x2  or  1/2  as  if  _      .        .   o      « 

the  equations  were  linear  equa-  21.         "^  »  ^  i'j 

tions  in  x2  and  y^.  X2/  =  2  x  -  2/  +  9. 

22    «2_x2/  +  2/2  =  37,  (x  +  2/)(8-x)  =  10, 

•  x2  -  j/2  =  40.  ■  (X  +  2/)  (5  -  2/)  =  20. 

24    (»2  +  2/2)(x  +  2/)=6,  25    (^  +  2/)^  =  3x2  -  2, 

'  «y(x  +  2/)  =  -  2.  *  •  (X  -  2/)2  =  32/2  -  11. 


SIMULTANEOUS  QUADRATIC  EQUATIONS 


117 


26. 

X  +  y/x^y  =  a, 
y  +  ^x?/2  =  6. 

28. 

X  +  y  =  58, 

Vx  +  Vy  =  10. 

30. 

X  +  2/  =  3. 

32. 

4x*-9y2  =  o, 

^    —O      ■       ..O             O  /            1       -A 

34. 


38. 


40. 


42. 


4x2  +  2/2  =  8(x  +  y). 
3x2/-2(x  +  y)  =  28, 
2x2/-3(x  +  2/)  =  2. 

x2  +  ?/2  +  X  +  y  =  18, 
x2  -  2/2  +  X  -  y  =  6. 

3x2- 2  2/2  =  6(x-2/), 
X2/  =  0. 

x2-X2/  +  2/2  =  13(x-2/), 
xy  =  12. 


Vx(l-y)  +V2/(l-ic)  =  a, 


2/ 


^^    Vl-x2Vl-2/2  +  X2/  =  |l, 
•  x-y  =  ll. 


X  _y  _  16 
46.  y      X  "  16' 

3x2  +  6y2  =  i20. 

X  V^  +  2/ Vy  _  1 
48.  xy/x  —  yy/y      2 


x3  -  8  =  8  -  y3. 


V2/  —  Va  —  X  =  Vy  -X, 
50.  Vy  —  X  +  Va  —  x  _  5 


52. 


Va  —  X 


1   ,   1_3 

i  +  i  =  l 
x2      y«      4 


27. 

29. 
31. 

33. 
35. 
37. 
39. 
41. 

43. 
45. 

47. 
49. 

51. 
53. 


xVx  +  y  =  3, 
y  Vx  +  y  =  1. 

■y/x  +y/y  =  a, 
X  +  y  =  6. 

Vx  —  Vy  =  2, 
(x  +  y)  Vxy  =  610. 


Vx-6  +  Vy  +  2  =  5, 
X  +  y  =  16. 
xy  +  xy-i  =  x2  +  y2, 
xy-xy-i  =  2(x2  +  y2). 

x-2  +  2  y-2  =  12, 

x-2  -  x-iy-i  +  y-2  =  4. 

5c2  +  y2_5(a;  +  y)^8, 
x2  +  y2-3(x  +  y)  =  28. 

2x2-3xy +  6y-  6  =  0, 
(x-2)(y-l)  =  0. 


V6-3x  +  x2  +  V6-3y  +  y2  =  6, 
X  +  y  =  3. 


X      y        ' 


y  =  .3. 


a     y  _  25 

y '^x~12* 
x2  -  y2  =  28. 


V5x  /x  + 

a;  +  y       \    6a 
icy  -  (x  +  y)  =  1. 

^  +  ?^  =  2, 
a2      62 

bx  +  ay  _m 

bx  —  ay      n 

x8  _  y3  _  16 
y        X        2  ' 

-  -1=-. 
y      x  ~  2' 


y  ^  3  V2 
X  2    ' 


Hint.  Let  -  =  u,  -  =  v. 

a        '  y 


118  QUADRATICS  AND  BEYOND 

X -1      a—1 


2, 
54.      )       ^i  55. 

3. 


y-\       b- 
x3  -  1      a3 


yS  _l  53  _  1 

lift  ^  +  ^  57  ^^^ 

^      2    x  +  l  221 

1      1_5  1_?-1 

5g    x'^y~6'  59    "^      ^~   ' 


X  +  22/  +  I2  x  + 

xy  =  2. 

PROBLEMS 

1.  Two  numbers  are  in  the  ratio  5 :  3.  Their  product  is  735.  What  are 
the  numbers  ? 

2.  Divide  the  number  100  into  two  parts  such  that  the  sum  of  their 

squares  is  5882. 

3.  The  sum  of  the  squares  of  two  numbers  increased  by  the  first  is 
205  ;  if  increased  by  the  second  the  result  is  200.    What  are  the  numbere  ? 

4.  The  diagonal  of  a  rectangle  is  85  feet  long.  If  each  side  were  longer 
by  2  feet,  the  area  would  be  increased  by  230  square  feet.  Find  the  length 
of  the  sides. 

5.  The  diagonal  of  a  rectangle  is  89  feet  long.  If  each  side  of  the  rec- 
tangle were  3  feet  shorter  the  diagonal  would  be  85  feet  long.  How  long 
are  the  sides  ? 

6.  The  sum  of  two  numbers  is  30.  If  one  decreases  the  first  by  3  and 
the  second  by  2  the  sum  of  the  reciprocals  of  the  diminished  numbers  is  ^. 
What  are  the  numbers  ? 

7.  The  sum  of  the  squares  of  two  numbers  is  370.  If  the  first  were 
increased  by  1  and  the  second  by  3,  the  sum  of  the  squares  would  be  500. 
What  are  the  numbers? 

8.  A  number  of  persons  stop  at  an  inn,  and  the  bill  for  the  entire  party 
is  $24.  If  there  had  been  3  more  in  the  party,  the  bill  would  have  been 
$33.    How  many  were  in  the  party  and  how  much  did  each  pay  ? 


SIMULTANEOUS  QUADRATIC  EQUATIONS  119 

9.  A  fruit  seller  gets  $2  for  his  stock  of  oranges.  If  his  stock  had  con- 
tained 20  more  and  he  had  charged  f  of  a  cent  more  for  each,  he  would  have 
received  $3  for  his  stock.  How  many  oranges  had  he  and  how  much  did  he 
get  apiece  for  them  ? 

10.  A  man  has  a  rectangular  plot  of  ground  whose  area  is  1250  square 
feet.  Its  length  is  twice  its  breadth.  He  wishes  to  divide  the  plot  into  a 
rectangular  flower  bed,  surrounded  by  a  path  of  uniform  breadth,  so  that 
the  bed  and  the  path  may  have  equal  areas.    Find  the  width  of  the  path. 

11.  In  going  7500  yards  one  of  the  front  wheels  of  a  carriage  makes  1000 
more  revolutions  than  one  of  the  rear  wheels.  If  the  wheels  were  each  a  yard 
greater  in  circumference,  the  front  wheel  would  make  625  more  revolutions 
than  the  rear  wheel.    What  is  the  circumference  of  the  wheels  ? 

12.  A  man  has  |539  to  spend  for  sheep.  He  wishes  to  keep  14  of  the 
flock  that  he  buys,  but  to  sell  the  remainder  at  a  gain  of  $2  per  head. 
This  he  does  and  gains  $28.  How  many  sheep  did  he  buy  and  at  what 
price  each? 

13.  A  man  buys  two  kinds  of  cloth,  brown  and  black.  The  brown  costs 
25  cents  a  yard  less  than  the  black,  and  he  gets  2  yards  less  of  it.  He 
spends  $28  for  the  black  cloth  and  $25  for  the  brown.  How  much  was  each 
a  yard  and  how  many  yards  of  each  did  he  get  ? 

14.  A  man  left  an  estate  of  $54,000  to  be  divided  among  8  persons,  namely, 
his  sons  and  his  nephews.  His  children  together  receive  twice  as  much  as 
his  nephews,  and  each  one  of  his  children  receives  $8400  more  than  each  one 
of  his  nephews.    How  many  sons  and  how  many  nephews  were  there  ? 

15.  A  and  B  buy  cloth.  B  gives  $9  more  for  60  yards  than  A  does  for 
45  yards ;  also  B  gets  one  yard  more  for  $9  than  A  ddes.  How  much  does 
each  pay? 

16.  A  sum  of  money  and  its  interest  amount  to  $22,781  at  the  end  of  a 
year.  If  the  sum  had  been  greater  by  $200  and  the  interest  :^  of  1  per 
cent  higher,  the  amount  at  the  end  of  the  year  would  have  been  $23,045. 
What  was  the  sum  of  money  and  what  was  the  interest  ? 

17.  If  one  divides  a  number  with  two  digits  by  the  product  of  its  digits, 
the  result  is  3.  Invert  the  order  of  the  digits  and  the  resulting  number  is  in 
the  ratio  7 :  4  to  the  original  number.    What  is  the  number  ? 

18.  What  number  of  two  digits  is  4  less  than  the  sum  of  the  squares  of 
its  digits  and  5  greater  than  twice  their  product  ? 

19.  Increase  the  numerator  of  a  fraction  by  6  and  diminish  the  denomi- 
nator by  2,  and  the  new  fraction  is  twice  as  great  as  the  original  fraction. 
Increase  the  numerator  by  3  and  decrease  the  denominator  by  the  same,  and 
the  fraction  goes  into  its  reciprocal.    What  is  the  fraction  ? 


120  QUADRATICS  AND  BEYOND 

121.  Equivalence  of  pairs  of  equations.  In  the  theorems  of 
this  section  the  capital  letters  represent  polynomials  in  x  and  y, 
and  the  small  letters  represent  numbers  not  equal  to  zero. 


I 


Theorem  I.    The  pairs  of  equations 
A 

are  equivalent. 

If  (ari,  2/1)  be  a  pair  of  values  that  satisfy  (1),  then  when  x  and 
y  in  5^  are  replaced  by  Xi  and  yi  the  equation  B"^  =  i^  is  a  numer- 
ical identity.  These  values  (xy,  y^  must  then  satisfy  one  of  the 
equations  -S  =  ±  ^,  for  if  they  did  not,  but  only  satisfied  the  equa- 
tion say  B  =  c  when  c  =^  ±b,  then  the  hypothesis  that  B^  =  }p-  is 
satisfied  by  (cci,  ?/i)  would  be  contradicted. 

Conversely^  any  pair  of  values  that  satisfy  B  =  -^h  evidently 
satisfy  B"-  =  b\ 

This  theorem  is  used,  for  instance,  in  exercise  2,  p.  114,  and  justifies  the 
assumption  that 

are  equivalent  pairs  of  equations. 

Theorem  II.    The  pairs  of  equations 

are  equivalent. 

li  A  =  a  and  B  =  h  are  satisfied  by  a  pair  of  numbers  (xj,  y^), 
we  multiply  the  identities  and  obtain  AB  =  ab. 

Conversely f  if  A  =  a,  AB  =  ab  are  identically  satisfied  by  a 
pair  (xi,  yi),  since  a  ^  0  we  can  divide  the  second  identity  by 
the  first  and  obtain  B  =  b.  Thus  if  (cci,  yi)  satisfy  one  pair  of 
equations  they  satisfy  the  other  pair. 

This  theorem  is  assumed  in  exercise  3,  p.  115,  to  show  that 

a;«  +  w«  =  12  ^  ,     a;2  -  XM  +  7/2  =  6 1  .     ,     ^ 

,  ^      y     and         .      "  ~r  ^  ^  are  equivalent. 


SIMULTANEOUS  QUADRATIC  EQUATIONS  121 


Theorem  III.    The  pairs  of  equations 


^:^}a)  ^f!^!:^K^) 


B  =  0]^  '     cA-\-dB  =  0 
are  equivalent  where  a,  b,  c,  and  d  are  numbers  such  that 

ad  —  be  ^  0. 

If  (xi,  yi)  satisfy  (1),  evidently  they  also  satisfy  (2).    Thus  all 
solutions  of  (1)  are  among  those  of  (2). 
Conversely  J  if  (xi,  ^/i)  satisfy  (2),  then 

__bB__dB 
a  c 

Thus  (ad  —  hc)B  =  0. 

Thus  since  (ad  —  bo)  =^  0,  . 

^  =  0. 

Similarly,  ^  =  0. 

This  theorem  has  been  assumed  in  exercises  1,  2,  3, 6,  p.  114.  In  1,  for  example, 
it  is  necessary  to  show  that 

are  equivalent.    In  this  case  a=c=l,  b  =  —  d  =  2.    Thus  ad  —  be  =  —  4:  ^t  0. 

122.  Incompatible  equations.  When  a  pair  of  simultaneous 
equations  can  be  proven  equivalent  to  a  pair  of  equations  which 
contradict  each  other  or  are  absurd,  they  are  incompatible  and 
have  no  finite  solution. 


(1) 
(2) 


Example  1. 

xy  =      1 

Subtract, 

xy  =  -1 
0=     2 

Example  2. 

x'^  +  y^  =  4, 
4x2  +  42/2  =  49. 

Multiply  (1)  by  4, 
Subtract, 

4x2  +  42/2  =  16 

4x2  +  42/2  =  49 

0  =  33 

122 


QUADRATICS  AND  BEYOND 


123.  Graphical  representation  of  simultaneous  quadratic  equa- 
tions. Every  equation  that  we  have  considered  may  be  rep- 
resented graphically  by  plotting 
in  accordance  with  the  method 
already  given  (p.  93). 

The  solution  of  simultaneous 
equations  is  represented  by  the 
points  of  intersection  of  the  cor- 
responding graphs. 

Thus  the  equations 

x'^,f  =  25, 
2xy  =  9 
have  the  solutions 


a;=±2± 


V34 


y==F2± 


V34 


or 


X  =  4.9,  .9,  -  .9,  -  4.9, 


y 


.9,  4.9,  -  4.9,  -  .9. 


These  equations  have  as  their  graph  the  preceding  jfigure. 
The  equations 


ic2  +  2£cH-4?/  +  l  =  0, 
x-f22/  +  4  =  0, 

which  have  the  solutions 

a;  =  ±  Vt  =  ±  2.6, 

y  =  -  2  =F  V^  =  -  3.3  or  - 


Y'^ 


-  ^  - 

^^ 

^N,                  X 

Z   '^ 

^J\             - 

/ 

^\r       - 

r 

%> 

7 

L^ 

!-—- 

5: 

r> 

~fcM — 

^^  ^ 

,       Ny 

1   /T 

"\    V 

I    ^ 

^  I^ 

-\    ^  ° 

-4      t* 

-^^^- 

J      t 

_    ^^ 

7 

"^ 

have  as  their  graph  the  figure  shown 
above. 

As  in  the  case  of  linear  equations, 
incompatible  equations  afford  graphs 
which  do  not  intersect.  Thus  the  graph 
of  the  equations  in  example  2,  p.  121,  is 
found  to  be  two  concentric  circles,  as 
is  shown  in  the  adjacent  figure. 


SIMULTANEOUS  QUADRATIC  EQUATIONS 


123 


Simultaneous  equations  which  have 
imaginary  solutions  also  lead  to  non-inter- 
secting graphs  (p.  101). 

Thus  the  equations 

a;2  +  2/'  =  4, 
have  the  adjacent  figure  as  their  graph. 


.  .^.- 

. 

^, 

V       : 

■      /"=VlXI    1    1 

1    >L  f 

0 

J      hs' 

^^^/|  1  1  U" 

EXERCISES 

1.  Inteipret  the  graphical  meaning  of  equivalent  pairs  of  equations. 

2.  Plot  and  solve  x^  +  ?/2  =  2, 

X  +  2/  =  2. 
What  general  statement  concerning  the  graphical  meaning  of  a  single 
solution  of  quadratic  and  linear  equations  does  this  example  suggest  ? 

3.  Plot  and  solve  the  following : 


(a) 


25. 


X2  +  2/2 

4x2  +  9^2  =  144. 


(b) 


(c) 


x2  +  2/2  =  25, 


x2  +  y2  =  25, 

4x2 -8x  + 92/2  =  140. 


5x2  +  2/2  =  25. 

What  general  statement  concerning  the  graphical  interpretation  of  four, 
three,  or  two  real  solutions  of  equations  do  these  examples  suggest  ? 

4.  State  the  algebraical  condition  under  which  two  quadratic  equations 
have  four,  three,  two,  or  one  real  solutions  (see  p.  113). 

5.  Plot  and  solve  the  following  : 
x2  +  2/  =  0,  „  .  a;2  +  2/2  =  9. 

32/ =  0. 


(a) 


(c) 


(e) 


X2 


4x-22/  =  3, 


^.2    I    2/2 
y^'  x2  -  y2  =  0. 


xy 


0. 


^    '    X2  +  y2 


16. 


4x2  +  92/2  =  36, 

x2  =  —  4  ?/. 


(f) 


xy  =  1, 
2x-Sy  =  lS. 


124.  Graphical  meaning  of  homogeneous  equations.  Consider  for  example 
the  homogeneous  equation 

3x2 -10x2/ -82/2  =  0.  (1) 


If  we  let  z  =  - ,  we  get 

X 


or 
or 


3  _  10  z  -  8  z2  =  0, 

8^2  +  102-3  =  0, 

(4z-l)(2z  +  3)  =0. 


124 


QUADRATICS  AND  BEYOND 


The  roots  are 

2  =  1    and    2  =  -  |. 

Thus 

y=i  and  y=-l, 

a;      4             X         2 

4y-x  =  0    and    3x  +  22/  =  0. 

or 

These  equations  represent  two  straight  lines  through  the  origin  which 
taken  together  form  the  graph  of  equation  (1).  This 
example  may  obviously  be  generalized :  Any  homo- 
geneous equation  of  the  form  ox^  +  hxy  +  cy"^  =  0  with 
positive  discriminant  represents  two  straight  lines 
through  the  origin.  Such  an  equation  is  equivalent 
to  two  linear  equations. 

In  an  example  like  5,  p.  115,  we  obtain  in  place 
of  the  given  pair  of  equations  a  pair  of  equivalent 
equations  one  of  which  is  homogeneous  and  the  other 
of  which  is  factorable.  We  can  learn  the  graphical 
meaning  of  tl;is  method  of  solution  by  studying  a 
particular  case.   Consider  for  example  the  equations : 

jc2  +  2x2/  +  7?/2  =  24,  (1) 

2x^-xy-y^  =  S.  (2) 

By  eliminating  the  constant  terms  we  obtain  the  product  of  the  two 
equations  x  +  y  =  0  and  x  —  2  y  =  0.  Thus  the  problem  of  solving  (1)  and 
(2)  is  replaced  by  that  of  solving  the  two  following  pairs  of  equations : 

x2  + 2X2/ +  72/2  =  24,  (^)      (2)  .  (2) 

x  +  y  =  0, 
or  x2  + 2x2/ +  73/2  =  24, 

X  -  2  2/  =  0. 

The  graphical  meaning  of  this 
method  of  solving  the  equations  (1) 
and  (2)  is  seen  in  the  fact  that  the 
problem  of  finding  the  points  of  inter- 
section of  the  graph  of  equation  (1) 
with  that  of  (2)  is  changed  to  that  of 
finding  the  intersection  of  the  graph  of  (1)  with  a  pair  of  straight  lines. 
This  appears  in  the  figure  where  the  curves  and  lines  are  numbered  as 
above.    The  closed  curve  represents  (1). 


CHAPTEE  XI 
MATHEMATICAL  INDUCTION 

125.  General  statement.  Many  theorems  are  capable  of  direct 
and  simple  proof  in  special  cases,  while  for  the  general  case  a 
direct  proof  is  difficult  and  complicated. 

If  we  ask  whether  ic"  —  1  is  divisible  by  x  —  1,  it  is  easy  to 
make  the  actual  division  for  any  particular  value  of  n,  as  n  =  2 
or  n  =  3.  But  if  x^  —  1  is  shown  divisible  by  a;  —  1,  we  are  no 
wiser  than  before  concerning  the  divisibility  of  x^  —  1.  Suppose, 
however,  we  can  prove  that  the  divisibility  for  ti  =  m  -f  1  follows 
from  that  for  n  =  m,  whatever  value  m  may  have.  Then  since 
we  have  established  the  fact  by  direct  division  for  n  =  3,  we  may 
be  assured  of  the  divisibility  for  ?i  =  4,  then  for  n  =  5,  and  so  on. 

Now  x'^+^  —  l=x(x"'  —  l)-\-(x  —  l) 

is  identically  true.  If  ic  —  1  is  a  factor  of  cc"*  —  1  for  a  given  value 
of  m,  it  is  a  factor  of  the  right-hand  member  and  consequently  a 
factor  of  the  left-hand  member  (§  69),  which  was  to  be  proved. 
Thus  the  divisibility  of  x""  —  1  hjx  —  1  is  established  for  any 
integral  value  of  n  greater  than  the  one  for  which  the  division 
has  actually  been  carried  out. 

To  complete  the  proof  of  a  theorem  by  mathematical  induction 
we  must  make  two  distinct  steps. 

^-^^^st    Establish  the  theorem  for  some  particular  case  or  cases, 
preferaUy  for  n  =1  and  n  =  2. 

Second.  Show  that  the  theorem  for  n  =  m  -{- 1  follows  from 
its  assumed  validity  for  n  =  m. 

Example.   Prove  that  the  sum  of  the  cubes  of  the  integers 

from  1  to  71  is  SK^(^  +  1)]S'- 

To  prove  that  1«  -f-  2»  4-  3*  +  •  •  •  -F  7i»  =  J^[7i(n  +  1)]^. 

125 


126  QUADRATICS  AND  BEYOND 

First.    This  theorem  is  true  for  n  =  1. 

For  1»  =  1  =  J  ^  [1  (2)]  p  =  12  =  1 

The  theorem  is  also  true  for  n  =  2. 

For    l»  +  2»  =  9=  J^[2(2  +  l)]P  =  G-6)2=32  =  9. 

Second.   Assume  the  theorem  for  n  =  m,* 

1«  +  2«  +  . . .  +  w«  =  J  J[m(m  +  1)]P. 

Add  (m  +  1)^  to  both  sides  of  the  equation, 

1«  +  2«  +  . .  •  +  m«  +  (m  +  1)«  =  ^[m(m  +  1)]  p  +  (m  +  1)* 

=  [(^m)2-|-m  +  l](m  +  l)2 


=  ( 4 )  (^  +  ly 


=  [i(mH-l)(m  +  2)]S 
which  is  the  form  desired,  i.e.  m  +  1  replaces  m  in  the  formula. 

EXERCISES 
Prove  by  mathematical  induction  that 

1.  1+3+6  +  .. .+(2n-l)=n2. 

2.  2  +  22  +  28  +  . . .  +  2"  =  2(2»  -  1). 

3.  3  +  6  +  9  +  ...  +3n  =  f n(n  +  l). 

4.  12  +  22  +  32  +  ...  +  n2  =  ^n(n  +  l)(2n  +  l). 

5.  13  +  28  +  33  +  . . .  +  n3  =  (1  +  2  +  3  +  . . .  +  n)2 

6.  42  +  72  +  102  +  . . .  +  (3  n  +  1)2  =  ^ n(6 n2  +  16n  +  11). 

7.  X"  —  y"  is  divisible  by  x  —  y  for  any  integral  values  of  n. 

8.  x2»  —  2/2"  is  divisible  by  x  +  y  for  any  integral  values  of  n. 

9.  1.2  +  2-3  +  3. 4  +  4.6  +  . ..  +  n.(n  + l)  =  ^n(n  +  l)  (n  + 2). 

10.  1  •  1  +  2  •  32  +  3  .  62  +  ...  +  n(2  n  -  1)2  =  ^n(n  +  l)(6n2  -  2n  -  1). 

11.  1.2.3+2.3.4+3.4.6  +  ...+n(n+l)(n+2)  =  in(?i+l)(n+2)(n+3). 

12.  (1«  +  28  +  38  +  • . .  +  n8)  +  8(16  +  26  +  35  +  . . .  +  n6) 

=  4(1  +  2  +  3+  ...  +n)8. 

*  This  statement  does  not  imply  that  we  assume  the  validity  of  the  formxila  for  any 
values  for  which  it  has  not  yet  been  established,  but  only  for  values  of  m  not  greater 
than  2. 


MATHEMATICAL  INDUCTION  127 


14. 

Ill                        1         _     ^ 

1.2'2.3'3.4'          'n-(w  +  l)      n  +  1 

15. 

o 

16. 

2.6  +  3.6  +  4.7  +  ...  +  («  +  l)(.  +  4)  =  "<»  +  ''g><''  +  «V 

17.  2.4  +  4.6  +  6.8  +  -..  +  2n(2n  +  2)  =  ^(2n  +  2)(2n  +  4) 

o 

18.  A  pyramid  of  shot  stands  on  a  triangular  base  having  m  shot  on  a 
side.    How  many  shot  are  in  the  pile  ? 


CHAPTEE  XII 

BINOMIAL  THEOREM 

126.  Statement  of  the  binomial  theorem.  When  in  previous 
problems  any  power  of  a  binomial  has  been  required  we  have 
obtained  the  result  by  direct  multiplication.  We  can,  however, 
deduce  a  general  law  known  as  the  binomial  theorem,  which  gives 
the  form  of  development  of  (a  +  hy,  where  n  is  any  positive  integer 
and  a  and  h  are  any  algebraical  or  arithmetical  expressions.  This 
law  is  as  follows : 

(a  4-  ^»)«  =  a«  +  ^  a«-  ^b  +  ^ '  ^f  ~  ^^  a^-^'b^  +  "-  +  b\ 
From  this  expression  we  deduce  the  following 
EULE  FOR  THE  DEVELOPMENT  OF  (a  +  bf. 

The  first  term  is  a". 
n 
1 

To  obtain  any  term  from  the  preceding  term,  decrease  the 
exponent  of  a  in  the  preceding  term  by  1  and  increase  the 
exponent  of  b  by  1  for  the  new  exponents.  Multiply  the  coefficient 
of  the  preceding  term  by  the  exponent  of  a,  and  divide  it  by  the 
exponent  of  b  increased  by  1  for  the  new  coefficient. 

Remark.  In  practice  it  is  usually  more  convenient  first  to  write  down  all  the 
terms  with  their  proper  exponents,  and  then  form  the  successive  coefficients. 

EXERCISES 

Verify  by  multiplication  the  rule  given  for  the  following  : 

1.  (a  +  6)».  2.  (x  -  2/)8. 

3.  (2a +  36)*.  4.  (v^+Vi^)'. 

5    (2  a -6)4.  6.  {x-y/y)\ 

7.  (3a -2  6)8.  8.  (a-ia;  +  6-iy)4. 

128 


BINOMIAL  THEOREM  129 

127.  Proof  of  the  binomial  theorem.  We  have  already  stated 
that  /         -j  \ 

{a  +  ^)«  =  a"  +  7  a--^b  +  ""^  ~  ^  a^-'b""  +  ••.  (1) 

and  have  seen  that  it  is  justified  for  every  particular  case  that 
we  have  tested.  By  complete  induction  we  now  prove  this 
theorem  when  ti  is  a  positive  integer. 

First.    Let  n  =  2. 

That  is,  (a  -{- by=  a^ -{- 2  ab  +  b\ 

This  expression  evidently  obeys  the  law  as  stated  in  (1). 
Second.   Assume  the  theorem  for  n  =  m. 

That  is,  (a  +  &)-  =  a-  4-  ^  a^-^h  +  '^^'^ -^)  ^m-2^2  +  ....  (2) 

Multiply  both  members  hj  a  -\-b, 

(a  4-  Z>)"'+i  =  a'«+i  ^ja'^h^j  a'^-^b'^ 

z 

Z 
This   expression   is   identical   with   (2)  except  that   (m  + 1) 
replaces  m.     Hence  the  theorem  is   established  so  far  as  the 
first  three  terms  are  concerned. 

128.  General  term.  Though  we  have  stated  the  binomial 
theorem  for  a>§eneral  value  of  ri,  we  have  only  established  the 
exact  form  of  the  first  three  terms. 

Let  {a  +  by  =  a'*  +  c^a^'-^b  +  Cga""  ^'^  ^ .  , 

We  note  that  the  sum  of  the  exponents  of  a  and  5  is  w  in  any 
term  of  the  development  of  {a  -\-  by.  Also  the  exponent  of  b  in 
the  (r  -\-  l)st  term  is  r. 

We  have  already  seen  that 

n  n(n  —  l) 

and  that  the  first  three  terms  are 

J.  ±  •  z 


130  QUADKATICS  AND  BEYOND 

respectively.    This  indicates  that  the  coefficient  of  the  next  term 

will  be  — ^^ — - — -^r-^ and  in  general  that  the  coefficient  of  the 

(r  +  l)st  term  has  the  form 

_n(n-l)---(n-r  +  l) 

l-2...r  '  W 

which  is  in  fact  the  form  that  our  rule  (§  126)  would  afford  for 
any  particular  value  of  r. 
This  affords  the  following 

Rule.    The  (r  +  l)st  term  of  {a  +  hf  is 

n{n-l)-"{n-r  +  l)  ^„_ ,  ^, 
1.2'-   r 

The  form  of  the  coefficient  may  be  easily  remembered  since  the 
denominator  consists  of  the  product  of  the  integers  from  1  to  r, 
while  the  numerator  contains  an  equal  number  of  factors  consist, 
ing  of  descending  integers  beginning  with  n. 

For  any  particular  values  of  n  and  r  we  could  easily  verify  the 
rule  by  direct  multiplication.    For  the  rigorous  proof  see  p.  178. 

EXERCISES 

Develop  by  the  binomial  theorem : 


/_a_  _  ^^Y 


Solution 


6-5-4-3  (_±y(_  VxV  6.5.4.3-2/  a  V/     VaV 

1.2.3.4VV^/V      a2/  l-2.3.4.6VVi/\      aV 
6. 5-4. 3.2.1/     V^y 
1.2.3.4.5.6V      a2/ 

x«        x2       X       a3       a6  a*   "^  a^a" 

2.  (f-fa)6.  3.  (Va  +  v'ft)'. 


BINOMIAL  THEOREM  131 

10.  (1  +  V^y  -  (1  -  V^)'.  11.  ( Vx  +  v^)*+  ( Vx  -  Vy)*. 

(2  X  3  v\^^ 
1 )  • 
Sy      2x7 

Solution :  n  =  10,  r  +  1  =  S. 

The  (r  +  l)st  term  of  (a  +  6)"  is  (§  128) 

n(n-l)-'-(n-r  +  1) 

1-2. -r 
In  this  case  we  get 

10-9-8.76-5.4    (^xy/Syy 

1.2.3.4-6-6.7  '\3y)\2x) 

33y3     27x7  24X* 

_  120-81    y^_  1215y4 
16      'x*~     2x*    ' 

(1\13 
a  +  -)   . 

14.  Find  the  6th  term  of  (—  -  ^V^. 

\2y       xj 

15.  Find  the  8th  term  of  ( —  -  ^  )   . 

\y  X  / 

16.  Find  the  6th  term  of  l2aVb )   . 

V  2aVb/ 

17.  Find  the  7th  term  of  (^  -  ^)   • 

18.  Find  correct  to  three  decimal  places  (.9)^ 
Solution;  (.9)8  =  (1  -  .1)8 

8^7^6^  ^ 

1.2.3.4^  /  V    '  -r 
=  1  -  8  •  0.1  +  28  •  0.01  -  66  .  0.001  +  70  •  0.0001 
=  1  +  0.28  +  0.0070  -  0.8  -  0.056 
=  1.2870  -  0.856  =  .431. 

In  this  exercise  any  terms  beyond  those  taken  would  not  affect  the  first 
three  places  in  the  result. 


132  QUADRATICS  AND  BEYOND 

Compute  the  following  correct  to  three  places : 

19.  (1.1)10.  20.  (2.9)8.  21.  (.98)11.  22.  (1.01)«^ 

23.  (1^)8.  24.  (1^)10.  25.  (98)8.  26.  (203)5. 

27.  In  what  term  of  (a  +  6)2o  does  a  term  involving  ai*  occur? 

28.  For  what  kind  of  exponent  may  a  and  h  enter  the  same  term  with 
equal  exponents  ? 

29.  For  what  kind  of  exponent  is  the  number  of  terms  in  the  binomial 
development  even  ? 

30.  Find  the  first  three  and  the  last  three  terms  in  the  development  of 

(Z, .  1      \24 


CHAPTEB  XIII 
ARITHMETICAL  PROGRESSION 

129.  Definitions.  A  series  of  numbers  such  that  each  numbei 
minus  the  preceding  one  always  gives  the  same  positive  or 
negative  number  is  called  an  arithmetical  series  or  arithmetical 
progression  (denoted  by  A.P.). 

The  constant  difference  between  any  term  and  the  preceding 
term  of  an  A.P.  is  called  the  common  difference. 

The  series  4,  7,  10,  13,  •  •  •  is  an  A.P.  with  the  common  difference  3.  The  series 
8,  62,  5,  32,  •  •  •  is  an  A.P.  with  the  common  difference  —  §.  The  series  4,  6,  7,  9, 
10,  •  •  •  is  not  an  A.P. 

EXERCISES 

Determine  whether  the  following  series  are  in  A.P.  If  so,  find  the  common 
difference. 

1.  6,  m  If,  ....  2.  27,  22^,  18,  .... 

3.  6,  4^  3,  H,  ....  4.  6,   -2,  -8,  .... 

5.  VI,   V2,  3  VI,  •••.  6.  8,  5|,  3|,  If,  .... 

-  1         2        4  V2-I     V2  1 

'•  V2     V2     V2  2  2       2(V2-1)  * 

9.3,   -^,   -3f,   -6f,  ....  10.^    ^^  +  2         -^^ 


6  6(V3-4) 

130.  The  nth.  term.  The  terms  of  an  A.P.  in  which  a  is  the 
first  term  and  d  the  common  difference  are  as  follows: 

a,  a -i- d,  a -\- 2d,  a -{- 3dj  '".  (1) 

The  multiple  of  d  is  seen  to  be  1  in  the  second  term,  2  in  the 
third  term,  and  in  fact  always  one  less  than  the  number  of  the 
term.    If  we  call  I  the  nth.  term,  we  have 

I  =  a  -{■  (^n  —  1)  d. 

133  ^  ..,    a 


134  QUADRATICS  AND  BEYOND 

We  may  also  write  the  series  in  which  I  is  the  nth  term  as* 

follows : 

a,  a-{-  dy  a  +  2d,  •  •  • ,  I  —  2d,  I  —  d,  I. 

131.  The  sum  of  the  series.  We  may  obtain  a  formula  for 
computing  the  sum  of  the  first  n  terms  of  an  A.P.  by  the  following 

Theorem.  The  sum  s  of  the  first  n  terms  of  the  series 
a,  a  -^dy  •  •  •,  I  —  d,  I  is 

By  definition, 

s  =  a-\-(a  +  d)  +  (a  +  2d)-\ -]-{l  -  2d)-\-(l  -  d)+ I.    (1) 

Inverting  the  order  of  the  terms  of  the  right-hand  member, 

s  =.  I  +  {l  -  d)  +  {I  -  2 d)  -^  '  ■  -  +  {a  +  2 d)  +  (a  -\-  d)+  a.    (2) 
Adding  (1)  and  (2)  term  by  term, 
2s^{l-\-a)+{l  +  a)-\-{l  +  a)  +  .'-^{l-\-a)  +  (l-\-a)  +  {l  +  a) 

z=n{a-\- 1). 
Thus  s  =  ^(a-\-l). 

132.  Arithmetical  means.  The  terms  of  an  A.P.  between  a 
given  term  and  a  subsequent  term  are  called  arithmetical  means 
between  those  terms.  By  the  arithmetical  mean  of  two  numbers 
is  meant  the  number  which  is  the  second  term  of  an  arithmetical 
series  of  which  they  are  the  first  and  third  terms.    Thus  the 

arithmetical  mean  of  two  numbers  a  and  h  is  — r— >  since  the 

numbers  a,  — ^r— >  h  are  in  A.P.  with  the  common  difference  — —  • 

The  two  formulas 

«  =  a  +  (»i  -  1)  d,  (I) 

«  =  -(«  + 1)  (II) 

contain  the  elements  a,  I,  s,  n,  d.  Evidently  when  any  three  are 
known  the  remaining  two  may  be  found  by  solving  the  two  equa- 
tions (I)  and  (II). 


AKITHMETICAL  PROGRESSION  135 

EXERCISES 

1.  Find  the  16th  term  and  the  sum  of  the  series  4,  2,  0,  —  2,  •  • .. 
Solution:  n  =  16,  a  =  4,  d  =  2  -  4  =  -  2. 

Z  =  a  +  (n  -  l)d  =  4  -t- 15(-  2)  =  -26,^ 

«  =  |(«4-0  =  f(4-26)=-176. 

2.  Z  =  42,  a  =  -  3,  d  =  3.    Find  n  and  s. 

3.  a  =  -  4,  n  =  8,  s  =  64.    Find  d  and  I. 

4.  d  =  -  i,  n  =  6,  «  =  21.    Find  s  and  a. 

5.  d  =  —  i,  n  =  10,  s  =  65.    Find  a  and  i. 

6.  s  =  161,  Z  =  4,  a  =  —  3.    Find  d  and  n. 

7.  Z  =  22,  s  =  243,  n  =  13.    Find  a  and  d. 

8.  s  =  -  15,  Z  =  -  2,  d  =  2.    Find  n  and  a. 

9.  d  =  41,  a  =  -  16,  s  =  140.    Find  n  and  i. 

10.  Insert  8  arithmetical  means  between  4  and  28. 

11.  Find  expressions  for  n  and  s  in  terms  of  a,  I,  and  d. 

12.  Find  expressions  for  I  and  a  in  terms  of  s,  n,  and  d. 

13.  Find  expressions  for  a  and  s  in  terms  of  d,  /,  and  n. 

14.  Find  expressions  for  d  and  n  in  terms  of  s,  a,  and  Z. 

15.  Find  the  13th  term  and  the  sum  of  the  series 

V2-1    V2  1 

2       '    2   '2(V2-l)'"" 

16.  Find  the  10th  term  and  the  sum  of  the  series 

V3    3V3  +  2    V3      2 
T"'~~6~"'"2"  +  3'"- 

17.  Insert  4  arithmetical  means  between  — -  and 

V2  2 

10\/6 


18.  Insert  6  arithmetical  means  between  -x  -  and 


4 


19.  Insert  3  arithmetical  means  between and 

2 

20.  Find  the  21st  term  and  the  sum  of  the  series ,   V^ 

V2 

V2 

21.  Find  the  10th  term  and  the  sum  of  the  series  — = 


136  QUADRATICS  AND  BEYOND 

22.  Eind  expressions  for  d  and  a  in  terms  of  s,  Z,  and  n. 

23.  Find  expressions  for  d  and  I  in  terms  of  a,  n,  and  s. 

24.  Find  the  8th  term  and  the  sum  of  the  series  x,  4ic,  7x,  •  •  •. 

25.  Find  the  9th  term  and  the  sum  of  the  series  8,  9J,  10|,  •  •  •. 

26.  Find  the  12th  term  and  the  sum  of  the  series  8,  7y\,  6|,  •  •  • . 

27.  Find  the  8th  term  and  the  sum  of  tlie  series  —  8,  —  4,  0,  •  •  •• 

28.  Find  the  12th  term  and  the  sum  of  the  series  27,  22 1,  18,  •  •  •. 

29.  Find  the  20th  term  and  the  sum  of  the  series  1,  —  2i,  —  6,  •  •  •. 

30.  Find  the  11th  term  and  the  sum  of  the  series  5,  —  3,  —  11,  •  •  •. 

31.  Find  the  9th  term  and  the  sum  of  tlie  series  x  —  y,  x,  x  -]-  y,  •  •  -.    * 

32.  Insert  n  —  2  arithmetical  means  between  a  and  I.  Write  the  first  tliree. 

Kemark.  Often  an  exercise  may  be  solved  more  simply  if  instead  of  assum- 
ing the  series  x,  x-\-  y,  z  +  2y,  ■■  -we  assume  x  —  y,  x,  x  -\-  y  when  three  terms 
are  required,  or  x  —  2y,  x  —  y,  x,  x  +  y,  x  +  2y  when  five  terms  are  required,  or 
x~-3y,x  —  y,x  +  y,x-i-3y  when  four  terms  are  required. 

33.  The  sum  of  the  first  three  terms  of  an  A. P.  is  15.  The  sum  of  their 
squares  is  83.    Find  the  sum  of  the  series  to  ten  terms. 

34.  Find  expressions  for  n  and  a  in  terms  of  s,  Z,  and  d.  For  what  real 
values  of  s,  Z,  and  d  does  a  series  with  real  terms  not  exist  ? 

35.  In  an  A.  P.  where  a  is  the  first  term  and  s  is  the  sum  of  the  first 
n  terms,  find  the  expression  for  the  sum  of  the  first  m  terms. 

36.  Find  expressions  for  n  and  I  in  terms  of  a,  s,  and  d.  For  what  real 
values  of  a,  s,  and  d  does  a  series  with  real  terms  not  exist  ? 

37.  If  each  term  of  the  series  (1),  §  130,  is  multiplied  by  m,  is  the  new 
series  in  A. P.,  and  if  so,  what  are  the  elements  of  the  new  series  ? 

38.  If  each  term  of  the  series  (1)  in  §  130  is  increased  by  6,  is  the  new 
series  in  A. P.,  and  if  so,  what  are  the  elements  of  the  new  series  ? 

39.  The  difference  between  the  third  and  sixth  terms  of  an  A.  P.  is  12. 
The  sum  of  the  first  10  terms  is  45.    Find  the  elements  of  the  series. 

40.  Find  the  10th  term  of  an  A. P.  whose  first  and  sixteenth  terms  are  3 
and  48.  Find  also  the  sum  of  those  eight  terms  of  the  series  the  last  of 
which  is  60. 

41.  Two  A.P.'s  have  the  same  common  difference,  and  their  first  terms 
are  2  and  4  respectively.  The  sum  of  the  first  seven  terms  of  one  is  to  the 
sum  of  the  first  seven  terms  of  the  other  as  4  is  to  6.  Find  the  elements  of 
both  series. 

42.  The  three  digits  of  a  number  are  in  A. P.  The  number  itself  divided 
by  the  sum  of  the  digits  is  48.  The  number  formed  by  the  same  digits  in 
reverse  order  is  396  less  than  the  original  number.    What  is  the  number? 


CHAPTEE  XIV 
GEOMETRICAL  PROGRESSION 

133.  Definitions.  A  series  of  nTimbers  such  that  the  quotient 
of  any  term  of  the  series  by  the  preceding  term  is  always  the 
same  is  called  a  geometrical  progression  (denoted  by  G.P.). 

The  constant  quotient  of  any  term  by  the  preceding  term  of  a 
G.P.  is  called  the  ratio. 

The  G.P.  series  4,  8, 16,  •  ■  •  has  the  ratio  2.   The  G.P.  series  8,  4  Vi,  4,  •  •  •  has 

V9 

the  ratio  J-^i . 

^'  EXERCISES 

Determine  which  of  the  following  series  are  in  G.P.  and  find  the  ratio. 
1.  4,  2,  1,  .... 
3.  8,  -2,  .5,  .... 

5.  Vl»  i»  Vl,  •••• 


2. 

4,  8,  16,  .... 

4. 

8,  -4,  -2,  .... 

6. 

6,-21,  73i,  ... 

8. 

^           2      ^ 

V2'       ''   V2' 

10. 

V3        [3"     V3 
8   '  V32'     4   ' 

7.-^,    A  2,  .... 

V2 

\  6     V5     Vl5 

11.  —J^ — -^,  5-2V6,3V3-V2,....    12.   V2  -  1,  1,  V2' +  1,  .... 

V3  -  V2 

134.  The  nth  term.  The  terms  of  a  G.P.  in  which  a  is  the 
first  term  and  r  the  ratio  are  as  follows : 

a,  ar,  ar^^  at^,  .... 

The  power  of  r  in  the  second  term  is  1,  in  the^  third  term  is 
2,  and  in  fact  is  always  one  less  than  the  number  of  the  term. 
If  we  call  I  the  nth  term,  we  have  the  following  expression  for 
the  T^th  term : 

137  ^ 


138  QUADRATICS  AND  BEYOND 

135.  The  sum  of  the  series.  We  obtain  a  formula  for  finding 
the  sum  of  the  first  7i  terms  of  a  G.P.  by  the  following 

Theorem.    The  sum  s  of  the  first  n  terms  of  the  geometrical 
progression  a,  ar,  ar^^  ...  is 

a  —  rl     .{■> 
'  =  T^r-     " 
By  definition,    s  =  a  -\-  ar  +  ar^  H-  •  •  •  4-  ar"~^ 
=  a(l  -h  r  +  r"^  -\ h  r^~^) 


K^) 


a  —  rar"^   ^      a  —  rl 


by  (III),  p.  15 
by  (I),  p.  137 


136.  Geometrical  means.  The  n  —  2  terms  between  the  first 
and  the  ?ith  term  of  a  G.P.  are  called  the  geometrical  means 
between  those  terms. 

If  one  geometrical  mean  is  inserted  between  two  numbers,  it 
is  called  the  geometrical  mean  of  those  numbers.  Thus  the 
geometrical  mean  between  a  and  h  is  ^ ah. 

The  two  fundamental  formulas 

«  =  ar»*-i,  (I) 

_  a{\-r^)  _  a-rl 

^  -    1-r    -  "rr7  ^^^^ 

contain  the  five  elements  a,  I,  r,  n,  s,  any  two  of  which  may  be 
found  if  the  remaining  three  are  given. 

EXERCISES 

1.  Find  the  7th  term  and  the  sum  of  the  G.P.  1,  4,  16,  . . .. 
Solution :  a  =  1,  n  =  7,  r  =  4. 

Substituting  in  (I),  I  =  ar*'-^  =  1  •  4^  =  4096, 

a  —  rl 


1-r 


o  V  *-^  *•       •    /TTv           1-4-4096      1638.3      ^,^, 
Substitutmg  m  (II),  s  = = =  6461. 


GEOMETRICAL  PROGRESSION  139 

2.  Insert  2  geometrical  means  between  4  and  32, 

3.  Insert  4  geometrical  means  between  32  and  1. 

4.  Insert  3  geometrical  means  between  3  and  |f . 

5.  Insert  4  geometrical  means  between  a^  and  l^. 

6.  Insert  4  geometrical  means  between  1  and  9V3. 

7.  Insert  3  geometrical  means  between  y-  a-iid  73|. 

8.  What  is  the  geometrical  mean  between  3  and  27  ? 

9.  Insert  3  geometrical  means  between  VS  and  "v/24. 

10.  Insert  4  geometrical  means  between  a  and  a^  Va6^. 

11.  Insert  3  geometrical  means  between  —  |  and  —  2^. 

12.  What  is  the  geometrical  mean  between  —  2  and  —  f  ? 

13.  Find  the  7th  term  and  the  sum  of  the  series  1,  3,  9,  •  •  •. 

14.  Find  the  6th  term  and  the  sum  of  the  series  2,  4,  8,  •  •  • . 

15.  What  is  the  geometrical  mean  between  -y/a^h  and  VoP  ? 

16.  Find  the  7th  term  and  the  sum  of  the  series  8,  2,  .5,  •  •  •. 

17.  Find  the  8th  term  and  the  sum  of  the  series  ^^j,  i^,  j,  •  •  • . 

18.  Find  the  7th  term  and  the  sum  of  the  series  Vi,  2,  2V2,  •  •  •. 

19.  Find  the  7th  term  and  the  sum  of  the  series  \/2,  V^,  Vi,  •  •  • . 

20.  Find  the  10th  term  and  the  sum  of  the  series  ■^\-^^  y^^,  ^^j,  •  •  •. 

21.  Find  the  5th  term  and  the  sum  of  the  series  V2  —  1,  1,  1  +  V2,  •  •  •. 

22.  The  first  and  sixth  terms  of  a  G.P.  are  1  and  243.    Find  the  interme- 
diate terms. 

23.  Find  the  5th  term  and  the  sum  of  the  series 
^        :,  6-2V6,  9V3-11V2,  ••.. 


V3+  Vii 

24.  Insert  3  geometrical  means  between  —  and  - . 

V3  9 

25.  What  is  the  geometrical  mean  between ~  and  ^^ ^? 

X  -\-  y  X  -y 

1  -     4 

26.  Find  the  6th  term  and  the  sum  of  the  series ,  —  •\/2,  — =,  •  • .. 

V2  V2 

27.  Find  the  6th  term  and  the  sum  of  the  series  -* /- ,  1,  — --,  •  •  •. 

\3         V2 

28.  Find  the  5th  term  and  the  sum  of  the  series  v^,  —  1, ,  •  •  •. 

29.  Find  the  6th  term  and  the  sum  of  the  series ,  -»  / — , ,  •  •  •. 

8       \32      4 


140  QUADRATICS  AND  BEYOND 

30.  The  geometrical  mean  of  two  numbers  is  4  and  their  sum  is  10.  Find 
the  numbers. 

31.  The  fourth  term  of  a  G.P.  is  192,  the  seventh  term  is  12,288.  Find 
the  first  term  and  the  ratio. 

32.  If  the  same  number  be  added  to  or  subtracted  from  each  tern^^^^f^ 
G.P.,  is  the  resulting  series  geometrical? 

33.  The  product  of  the  first  and  last  of  four  numbers  in  G.P.  is  64. 
Their  quotient  is  also  64.    Find  the  numbers. 

34.  The  product  of  four  numbers  in  G.P.  is  81.  The  sum  of  the  second 
and  third  terms  is  i.    Find  the  numbers. 

35.  If  every  term  of  a  G.P;  be  multiplied  by  the  same  number  m,  is  the 
resulting  series  a  G.P.?    If  so,  w^hat  are  the  elements? 

36.  The  sum  of  three  numbers  in  G.P.  is  42.  The  difference  between  the 
squares  of  the  first  and  the  second  is  60.    What  are  the  numbers  ? 

37.  The  difference  between  two  numbers  is  48.  The  arithmetical  mean 
exceeds  the  geometrical  mean  by  18.    Find  the  numbers. 

38.  Four  numbers  are  in  G.P.  The  difference  between  the  first  and  the 
second  is  4,  the  difference  between  the  third  and  the  fourth  is  36.  Find  the 
numbers. 

39.  A  ball  falling  from  a  height  of  60  feet  rebounds  after  each  fall  one 
third  of  the  last  descent.  What  distance  has  it  passed  over  when  it  strikes 
the  ground  for  the  eighth  time  ? 

40.  The  difference  between  the  first  and  the  last  of  three  terms  in  G.P. 
is  four  times  the  difference  between  the  first  and  second  terms.  The  sum  of 
the  numbers  is  208.    Find  the  numbers. 

41.  An  invalid  on  a  certain  day  was  able  to  take  a  single  step  of  18 
inches.  If  he  was  each  day  to  walk  twice  as  far  as  on  the  preceding  day, 
how  long  before  he  can  take  a  five-mile  walk  ? 

42.  The  difference  between  the  first  and  the  last  of  four  numbers  in  G.P. 
is  thirteen  times  the  difference  between  the  second  and  third  terms.  The 
product  of  the  second  and  third  terms  is  3.    Find  the  numbers. 

137.  Infinite  series.  When  the  number  of  terms  of  a  G.P.  is 
unlimited  it  is  called  an  infinite  geometrical  series. 

In  the  series  a,  ar,  ar^^  •  •  •,  when  r  >  1,  evidently  each  term  is 
larger  than  the  preceding  term.  The  series  is  then  called  increas- 
ing. When  r  <  1,  each  term  is  smaller  than  the  preceding  term 
and  the  series  is  called  decreasing. 

T,T        .  a  (1  —  r")  a  ar^ 

Now  m  any  case  s  =  — \ = 

1  —  r  1  —  r       1  —  r 


GEOMETRICAL  PROGRESSION  141 

When  r  >  1,  evidently  r"  becomes  very  large  for  large  values 
of  n.  For  this  case,  then,  the  sum  of  the  first  n  terms  becomes  very 
large  for  large  values  of  n.  In  fact  we  can  take  enough  terms 
so  that  s  will  exceed  any  number  we  may  choose.  If,  however, 
r  <  1,  as  71  increases  in  value  r"  becomes  smaller  and  smaller.  In 
fact  we  can  choose  n  large  enough  so  that  r"  is  as  small  as  we  wish, 
or  as  we  say,  approaches  0  as  a  limit.  But  since  r"  may  be  made 
as  small  as  we  wish,  ar^  also  approaches  0  as  a  limit,  and  conse- 

quently approaches  0  as  a  limit.    Thus  when  r  <  1  the 

value  of  the  sum  of  the  first  n  terms  approaches as  n 

1  —  r 

becomes  very  great.    This  we  express  in  other  words  by  asserting 
that  the  sum  of  the  infinite  series 

a  -\-  ar  -\-  ar^  +  ■  "j  when  r  <  1, 

is  s^  = 


1-r 

.    EXERCISES 

Find  the  sum  of  the  following  infinite  series. 

1.  6  +  3  +  1  +  .... 
Solution :  a  =z  6,     r  =  |. 

a 


1-r 


6 

=  5  =  12. 

i-i 

h 

3. 

64  +  8  +  1+.. 

5. 

^  +  i^  +  :rV  +  - 

7. 

2 +  .5 +  .125  + 

2.  l  +  i  +  i  +  .... 
4.  h  +  \  +  l  +  -"' 
6.  !  +  f  +  -V  +  --- 

■v/2 
8.  V2  +  l  +  -y-  +  ....  9.  (V2  +  l)  +  l  +  (v^-l)  +  .... 

10.  How  large  a  value  of  n  must  one  take  so  that  the  sum  of  the  first 
n  terms  of  the  following  series  differs  from  the  sum  to  infinity  by  not  more 
than  .001? 

(a)  8  +  4  +  2  +  .... 

Solution :  a  =  8,     r  =  |. 

a  ar'*  ar** 

8  = =  Sao  — 


1-r      1-r  1 


8^-8= 

1-r 


142 


QUADRATICS  AND  BEYOND 


We  must  find  for  what  value  of  n  the  expression 


is  less  than  .001. 


say 


8-2 

2» 


16 

2«* 


'"^        -          -         16         1  ^ 

By  trial  we  see  that  if  w  =  14  the  value  of  —  is ,  which  is  less  than 

.001  2".    1^2^ 

(b)  27  +  3  +  i  +  . . ..  (c)  4  +  I  +  ^ij  +  •  •  •. 

(d)  1  +  ^1^  +  3,1^  +  ....  (e)  64  +  16  +  4  +  .... 

(f)  100  +  20  +  4  +  .  .  •.  (g)  60  +  20  +  6f%  . . .. 

11.  What  is  the  value  of  the  following  recurring  decimal  fractions? 

(a)  .212121.... 

Solution :  This  decimal  may  be  written  in  the  form 


Here 


(b)  .333.... 
(e)  .343343 


21           21             21 

100  '  (100)2  '  (100)3"'"     • 

21               1- 
100 '          100 

5    _     «     _     .21     _  .21  _  7 
*      1-r      l-.Ol      .99      33* 

(c)  .717171.... 

(d)  .801801.. 

(f)  1.43131....      ^ 

(g)  2.61414.. 

ADVANCED  ALGEBRA     • 

CHAPTEE  XV 
PERMUTATIONS  AND  COMBINATIONS 

138.  Introduction.  Before  dealing  directly  with  the  subject  of 
the  chapter  we  must  answer  the  question,  In  how  many  distinct 
ways  may  two  successive  acts  be  performed  if  the  first  may  be 
performed  in  p  ways  and  the  second  may  be  performed  in  q  ways  ? 
Suppose  for  example  that  I  can  leave  a  certain  house  by  any 
one  of  four  doors,  and  can  enter  another  house  by  any  one  of  five 
doors,  in  how  many  ways  can  I  pass  from  one  house  to  the  other  ? 
If  I  leave  the  first  house  by  a  certain  door,  I  have  the  choice  of 
all  five  doors  by  which  to  enter  the  second  house.  Since,  how- 
ever, I  might  have  left  the  first  by  any  one  of  its  four  doors, 
there  are  4  •  5  =  20  ways  in  which  I  may  pass  from  one  house 
to  the  other.    This  leads  to  the  v,^ 

Theorem.  If  a  certain  act  may  he  'performed  in  p  ways,  and 
if  after  this  act  is  performed  a  second  act  may  he  performed  in  q 
ways,  then  the  total  numher  of  ways  in  which  the  two  acts  may 
he  performed  is  p  •  q. 

With  each  of  the  p  possible  ways  of  performing  the  first  act 
correspond  q  ways  of  performing  the  second  act.  Thus  with  all 
the  p  possible  ways  of  performing  the  first  act  must  correspond 
q  times  as  many  ways  of  performing  the  second  act.  That  is,  the 
two  acts  may  be  performed  in  ^  •  5'  ways. 

It  is  of  course  assumed  in  this  theorem  that  the  performance 
of  the  second  act  is  entirely  independent  of  the  way  in  which  the 
first  act  is  performed. 

143 


144  ADVANCED  ALGEBRA 

EXERCISES 

1.  I  have  four  coats  and  five  hats.    How  many  different  combinations 
coat  and  hat  can  I  wear  ? 

Solution :  The  first  act  consists  in  putting  on  one  of  my  coats,  whicli  may 
be  done  in  four  ways-,  the  second  act  consists  in  putting  on  one  of  my 
hats,  which  may  be  done  in  five  ways.  Thus  I  have  4  •  5  =  20  different 
combinations  of  coat  and  hat. 

2.  In  how  many  ways  may  the  two  children  of  a  family  be  assigned  to 
five  rooms  if  they  each  occupy  a  separate  room  ? 

3.  A  gentleman  has  four  coats,  six  vests,  and  eight  pairs  of  trousers.  In 
how  many  different  ways  can  he  dress  ? 

4.  I  can  sail  across  a  lake  in  any  one  of  four  sailboats  and  row  back 
in  any  one  of  fifteen  rowboats.    In  how  many  ways  can  I  make  the  trip  ? 

5.  Two  men  wish  to  stop  at  a  town  where  there  are  six  hotels  but  do  not 
wish  quarters  at  the  same  hotel.    In  how  many  ways  may  they  select  hotels  ? 

6.  A  man  is  to  sail  for  England  on  a  steamship  line  that  runs  ten  boats 
on  the  route,  and  return  on  a  line  that  runs  only  six.  In  how  many  different 
ways  can  he  make  the  trip  ? 

7.  In  walking  from  A  to  B  one  may  follow  any  one  of  three  roads;  in 
going  on  from  B  to  C  one  has  a  choice  of  five  roads.  In  how  many  different 
ways  can  one  walk  from  A  to  C  ? 

139.  Permutations.  Each  different  arrangement  either  of  all 
or  of  a  part  of  a  number  of  things  is  called  a  permutation. 

Thus  the  digits  1,  2  have  two  possible  permutations,  taken  both 
at  a  time,  namely,  12  and  21. 

The  digits  1,  2,  3  have  six  different  permutations  when  two  are 
taken  at  a  time,  namely,  12,  13,  21,  23,  31,  32.  For  if  we  take  1 
for  the  first  place,  we  have  a  choice  of  2  and  3  for  the  second 
place,  and  we  get  12  and  13.  If  2  is  in  the  first  place,  we  get  21 
and  23.  Similarly,  we  get  31  and  32.  In  this  process  it  is  noted 
that  we  can  till  the  first  of  the  two  places  in  any  one  of  three 
ways ;  the  second  place  can  be  filled  in  each  case  in  only  two  ways. 
Thus  by  the  Theorem,  §  138,  we  should  expect  3-2  =  6  permuta- 
tions of  three  things  taken  two  at  a  time.  We  observe  that  this 
product  3  •  2  has  as  its  first  factor  3,  which  is  the  total  number  of 
things  considered.  The  number  of  factors  is  equal  to  the  number 
of  digits  taken  at  a  time,  i.e.  two.    This  leads  to  the  general 


PEKMUTATIONS  AND  COMBINATIONS  145 

Theorem.    Tlie  number  of  permutations  of  n  ohjects  taken  r 

at  a  time  is  ,         i\       ,  ,    -t\  /t\ 

n(n  —  1)  ■  ■ '  (n  —  r  -\-  J^)'  v-j 

This  is  symbolized  by  P„^  ^. 

This  formula  is  easily  remembered  if  one  observes  that  the  first 
factor  is  7i,  the  total  number  of  objects  considered,  and  that  the 
number  of  factors  is  r,  the  number  of  objects  taken  at  a  time. 
Thus  ■Py3  =  7-6-5. 

We  prove  this  theorem  by  complete  induction. 

Flrstj  let  r  =  1.  There  are  evidently  only  n  different  arrange- 
ment of  n  objects,  taking  one  object  at  a  time,  namely  (assuming 
our  objects  to  be  the  first  7i  integers), 

1,  2,  3,  ..-,  71. 

Let  us  take  two  objects  at  a  time,  i.e.  let  r=  2.    Since  there 

are  n  objects,  we  have  n  ways  of  filling  the  first  of  the  two  places. 

When  that  is  tilled  there  are  n  —  1  objects  left,  and  any  one  may 

be  used  to  fill  the  second  place.    Thus,  by  the  Theorem,  §  138, 

there  are  for  r  =  2  .         . . 

n  {n  —  V) 

different  permutations. 

Secondy  assume  the  form  (I)  for  r  =  m^ 

Pn,m  =  n(n-l)..-(n-m  +  l).  (1) 

We  can  fill  the  first  m  places  in  P„  ^  different  ways  since  there 
are  that  number  of  permutations  of  n  things  taken  m  at 'a  time. 
This  constitutes  the  first  act  (§  138).  The  second  act  consists  in 
filling  the  m  -f  1st  place,  which  may  be  done  in  n  —  m  ways  by 
using  any  of  the  remaining  n  —  in  objects.  Thus  the  number  of 
permutations  of  n  things  taken  7n  +  1  at  a  time  is 
Pn,r,^  +  i  =  Pn,nr(n-m)=:=n(n-l){n-2)---(n-m  +  l)(n-m\ 
which  is  the  form  that  (1)  assumes  on  replacing  m  by  m  +  1. 

Corollary.    The  number  of  permutations  of  n  things  taken  all 

at  a  time  is  ^  ,        -,\       ^   -t         » *  /o\ 

F^^^=n(n-'l)-'-2-l  =  n!*  (^) 

Taking  n  =  r  in  (I),  we  get  (2). 

»  « /  is  the  symbol  for  1  •  2  .3  •  4  •••(»-  1)  n,  and  is  read  factorial  n. 


146  ADVANCED  ALGEBRA 

EXERCISES 

1.  How  many  permutations  may  be  formed  from  8  letters  taken  four  at  a 
time? 

Solution :  n  =  8,  r=»4,  n-r  +  l=6, 

Pg,  4  =  8  •  7  .  6  .  5  =  1680. 

2.  In  how  many  different  orders  may  6  boys  stand  in  a  row  ? 

3.  How  many  different  numbers  less  than  1000  can  be  formed  from  the 
digits  1,  2,  3,  4,  5  without  repetition  ? 

4.  How  many  arrangements  of  the  letters  of  the  alphabet  can  be  made 
taking  three  at  a  time  ? 

5.  How  many  numbers  between  100  and  10,000  can  be  formed  from  the 
digits  1,  2,  3,  4,  5,  6  without  repetition  ? 

6.  How  many  different  permutations  can  be  made  of  the  letters  in  the 
word  compute  taking  four  at  a  time  ? 

7.  In  a  certain  class  there  are  4  boys  and  5  girls.  In  how  many  orders  may 
they  sit  provided  all  the  boys  sit  on  one  bench  and  all  the  girls  on  another  ? 

Hint.  Use  Corollary  §  139,  and  then  Theorem,  §  138. 

8.  I  have  6  books  with  red  binding  and  3  with  brown.  In  how  many  ways 
may  I  arrange  them  on  a  shelf  so  that  all  the  books  of  one  color  are  together  ? 

140.  Combinations.  Any  group  of  things  that  is  independent  of 
the  order  of  the  constituents  of  the  group  is  called  a  combination. 

The  committee  of  men  Jones,  Smith,  and  Jackson  is  the  same 
as  the  committee  Jackson,  Jones,  and  Smith.  The  sound  made  by 
striking  simultaneously  the  keys  EGrC  of  a  piano  is  the  same  as 
the  sound  made  by  striking  CGE.  In  general  a  question  involv- 
ing the  number  of  groups  of  objects  that  may  be  formed  where 
the  character  of  any  group  is  unaltered  by  any  change  of  order 
among  its  constituent  parts  is  a  question  in  combinations. 

Suppose  for  example  that  we  ask  how  many  committees  of  three 
men  can  be  selected  from  six  men.  If  the  men  are  called  A,  B,  C, 
D,  E,  F,  there  are,  by  §  139,  6  -5  •  4  =  120  difPerent  arrangements 
or  permutations  of  the  six  men  in  groups  of  three.  But  the  permu- 
tations A,  B,  C  ;  A,  C,  fe  ;  B,  A,  C,  etc.  (3 !  =  6  in  all  for  the  men 
A,  B,  and  C),  are  all  distinct,  while  evidently  the  six  committees 
consisting  of  A,  B,  and  C  are  identical.  This  is  true  for  every 
distinct  set  of  three  men  that  we  could  select;  that  is,  for  the 


PERMUTATIONS  AND  COMBINATIONS  147 

six  different  permutations  of  any  three  men  there  is  only  one 
distinct  committee.    Hence  the  number  of  committees  is  one  sixth 

the  total  number  of  permutations,  or  -^' 
This  leads  to  the  general 

Theorem.    The  number  of  comhinations  of  n  things  taken  r  at 

a  time  is  .        -,^       ,  .    ^x 

n(n  — I)--- (n —  T-\- 1) 

This  is  symbolized  by  C„^  ^. 

The  number  of  permutations  of  n  things  taken  r  at  a  time  is 

p^^^  =  n(n  -  1)  • '  •  (r,  —  r  +  1). 

In  every  group  of  r  things  which  form  a  single  combination 
there  are  (Cor.,  p.  145)  r !  permutations.  Thus  there  are  r !  times 
as  many  permutations  as  combinations.    That  is, 

r      -?Ji^-  n{n-l)-'-{n-r-\-l) 
"''•""/•!"  r\  '  ^^ 

This  formula  is  easily  remembered  if  one  observes  that  there 
is  the  same  number  of  factors  in  the  numerator  as  in  the  denomi- 
nator.   Thus 

10- 9- 8 
^10.3-  -^.2-3 
Corollary.    C  ^=  C 


n,  n  —  r 


Multiplying  numerator  and  denominator  of  (I)  by  (n  —  r)!, 
_n(n-l)-'-(n-r  +  l)(n-r)-'-2-l 
"•'•"  rl(n-ry. 


_n{n-l)---{r-\-l) 
(n  —  r)\ 

—  ^  (^  ~  1)  •  •  •  [^  ~  (^  ~  ^)  +  ^] 

(n  —  r)\ 


n.n  —  r' 


This  corollary  saves  computation  in  some  cases.  For  instance,  if  we  wish  to 
compute  Ci9,  ir,  it  is  more  convenient  to  write  Ci9, 17=  Ci9, 2  =  ^  ,  =  171  than  the 
expression  for  C'lo,  17. 


148  ADVANCED  ALGEBRA 

EXERCISES 

1.  How  many  committees  of  5  men  can  be  selected  from  a  body  of  10 
men  three  of  whom  can  serve  as  chaii'man  but  can  serve  in  no  other  capacity  ? 

Solution :  There  are  7  men  who  may  fill  4  places  on  the  committee. 


^'        1.2.3-4 


There  are  3  men  to  select  from  for  the  remaining  place  of  chairman, 
and  the  selection  may  be  made  in  3  ways.  Thus  the  committee  can  be 
made  up  in  3  •  35  =  105  ways. 

2.  How  many  distinct  crews  of  8  men  may  be  selected  from  a  squad  of 
14  men  ? 

3.  How  many  distinct  triangles  can  be  drawn  having  their  vertices  in 
10  given  points  no  three  of  which  are  in  a  straight  line  ? 

4.  How  many  distinct  sounds  may  be  produced  on  9  keys  of  a  piano  by 
striking  4  at  a  time  ? 

5.  In  how  many  ways  can  a  crew  of  8  men  and  a  hockey  team  of  5  men 
be  made  up  from  20  men  ? 

6.  In  how  many  ways  may  the  product  a-b  •  c  •  d-  e  -f  be  broken  up 
into  factors  each  of  which  contains  two  letters? 

7.  If  8  points  lie  in  a  plane  but  no  three  in  a  straight  line,  how  many 
straight  lines  can  be  drawn  joining  them  in  pairs? 

8.  How  many  straight  lines  can  be  drawn  through  n  points  taken  in 
pairs  no  three  of  which  are  in  the  same  straight  line  ? 

9.  Seven  boys  are  walking  and  approach  a  fork  in  the  road.  They 
agree  that  4  shall  turn  to  the  right  and  the  remainder  turn  to  the  left.  In 
how  many  ways  could  they  break  up  ? 

Solution :  The  number  of  groups  of  4  boys  that  can  be  formed  from  the 

CV  4  = =  o5. 

4! 

For  each  group  of  4  boys  there  remains  only  a  single  group  of  3  boys. 
Thus  the  total  number  of  ways  in  which  tliQ  party  can  divide  up  is 
precisely  35. 

10.  If  there  are  12  points  in  space  but  no  four  in  the  same  plane,  how 
many  distinct  planes  can  be  determined  by  the  points? 

Hint.  Three  points  determine  a  plane. 

11.  Eight  gentlemen  meet  at  a  party  and  each  wishes  to  shake  hands 
Tfith  all  the  rest.    How  many  hand  shakes  are  exchanged  ? 


I 


PERMUTATIONS  AND  COMBINATIONS  149 

12.  In  how  many  ways  can  a  baseball  team  of  9  men  be  selected  from 
14  men  only  two  of  whom  can  pitch  but  can  play  in  no  other  position  ? 

13.  How  many  baseball  teams  can  be  selected  from  15  men  only  four  of 
whom  can  pitch  or  catch,  provided  these  four  can  play  in  either  of  the  two 
positions  but  cannot  play  elsewhere  ? 

14.  Two  dormitories,  one  having  3  doors,  the  other  having  6  doors,  stand 
facing  each  other.  A  path  runs  from  each  door  of  one  to  every  door  of  the 
other.    How  many  paths  are  there  ? 

15.  Show  that  the  number  of  ways  in  which  p  -{-  q  things  may  be  divided 

into  groups  of  p  and  q  things  respectively  is  ^ ^^• 

p\q\ 

16.  Out  of  8  consonants  and  3  vowels  how  many  words  can  be  formed 
each  containing  3  consonants  and  2  vowels  ? 

17.  A  boat's  crew  consists  of  8  men,  three  of  whom  can  row  only  on  one 
side  and  two  only  on  the  other.    In  how  many  ways  can  the  crew  be  arranged  ? 

18.  A  pack  of  cards  contains  62  distinct  cards.  In  how  many  different 
ways  can  it  be  divided  into  4  hands  of  13  cards  each  ? 

19.  Five  points  lie  in  a  plane,  but  no  three  in  any  other  plane.  How 
many  tetrahedrons  can  be  formed  with  these  points  taken  with  two  points 
not  in  the  plane  ? 

141.  Circular  permutations.  By  circular  permutations  we 
mean  the  various  arrangements  of  a  group  of  things  around  a 
circle. 

Theorem.  The  number  of  orders  in  which  n  things  may  he 
arranged  in  a  circle  is  (n—l) !. 

Suppose  A  is  at  the  point  at  which  we  begin  to  arrange  the 
digits  1,  2,  3,  •  •  •,  n.  Suppose  we  start  our  arrangement  of  digits 
at  A  with  a  given  digit  a.  We  have  then 
virtually  n—l  places  to  fill  by  the  remaining 
n—l  digits.  Thus  we  get  {n—l)\  (p.  145) 
permutations  of  the  n  digits  keeping  a  fixed. 
But  suppose  we  start  our  arrangement,  that  is, 
fill  the  place  at  A  with  any  other  digit,  as  h, 
and  the  remaining  places  in  any  order  what- 
ever. If  we  now  go  around  the  circle  till  we 
come  to  the  digit  a,  the  succession  of  digits  from  that  point 
around  the  circle  to  a  again  must  be  one  of  the  {n  —1)\  orders 


150  ADVANCED  ALGEBRA 

which  we  obtained  when  we  took  a  as  the  initial  figure.  Thus 
the  only  distinct  orders  in  which  the  n  digits  can  be  arranged 
on  a  circle  are  the  (n  —1)\  permutations  we  obtained  by  filling 
the  first  place  with  a. 


EXERCISES 

1.  In  how  many  orders  can  6  men  sit  around  a  circular  table  ? 
Solution : 


n  =  Q,        n  -  1  =  5,         (n  -  1)!  =  5  !  =  120. 

2.  In  how  many  ways  can  8  men  sit  around  a  circular  table  ? 

3.  In  how  many  ways  may  the  letters  of  live  be  arranged  on  a  circle  ? 

4.  In  how  many  ways  may  the  letters  of  permutation  be  arranged  on  a 
circle  ? 

5.  In  how  many  ways  can  4  men  and  4  ladies  sit  around  a  table  so  that 
a  lady  is  always  between  two  men  ? 

6.  In  how  many  ways  may  4  men  and  their  wives  be  seated  around  a  table 
so  that  no  man  sits  next  his  wife  but  the  men  and  the  women  sit  alternately  ? 

7.  In  how  many  ways  can  six  men  and  their  wives  be  seated  around  a 
table  so  that  each  man  sits  between  his  wife  and  another  lady  ? 

8.  In  how  many  ways  can  10  red  flowers  and  5  white  ones  be  planted 
around  a  circular  plot  so  that  two  and  only  two  red  ones  are  adjacent  ? 

142.  Theorem.    The  number  of  permutations  of  n  things  of 

.    n! 
which  p  are  alike,  taken  all  together y  is  -^• 

If  all  the  things  were  different,  we  should  have  n !  permutations. 
But  since  p  of  the  n  things  are  alike,  any  rearrangement  of  those 
p  like  things  will  not  change  the  permutation.  Eor  any  fixed 
arrangement  of  the  n  things  there  are  p !  different  arrangements 

of  the  p  like  things.    Thus  — :  of  the  n\  permutations  are  iden- 

tical,  and  there  are  only  — '-  distinct  permutations  of  the  n  things 
^p  of  which  are  alike.  ' 


Corollary.    If  of  n  things  p  are  of  one  kind,  q  of  another 

n  f 
kind,  r  of  another,  etc.,  then  there  are  — ; — -^— —  permutations 

of  the  n  things  taken  all  at  a  time.       -^ '  ^'     '     ^ 


PERMUTATIONS  AND  COMBINATIONS  151 

EXERCISES 

1.  How  many  distinct  arrangements  of  the  letters  of  the  word  Cincinnati 
are  possible  ? 

Solution  :  There  are  in  all  10  letters,  of  which  3  are  i,  2  are  c,  and  3  are  n. 
Thus  the  number  of  arrangements  is 

10!      _X.2.^-^-5-^-7-8-9.10 
3!3!2!"        ;.^.3X-^-^;-;Z 
=  2.5-7.8.9.10  =  50,400. 

2.  How  many  distinct  arrangements  of  the  letters  of  the  word  parallel 
can  be  formed  ? 

3.  How  many  signals  can  be  made  by  hanging  15  flags  on  a  staff  if  2 
flags  are  white,  3  black,  5  blue,  and  the  rest  red  ? 

4.  How  many  signals  can  be  made  by  the  flags  in  exercise  3  if  a  white 
one  is  at  each  extreme  ? 

5.  How  many  signals  can  be  made  by  the  flags  in  exercise  3  if  a  red  flag 
is  always  at  the  top  ? 

6.  Would  3  dots,  2  dashes,  and  1  pause  be  enough  telegraphic  symbols 
for  the  letters  of  the  English  alphabet,  the  numerals,  and  six  punctuation 
marks  ? 


CHAPTEE  XVI 
COMPLEX  NUMBERS 

143.  The  imaginary  unit.  When  we  approached  the  solution 
of  quadratic  equations  (p.  52)  we  saw  that  the  equation  x^  =  2 
was  not  solvable  if  we  were  at  liberty  to  use  only  rational  num- 
bers, but  that  we  must  introduce  an  entirely  new  kind  of  number, 
defined  as  a  sequence  of  rational  numbers,  if  we  wished  to  solve 
this  equation.  The  excuse  for  introducing  such  numbers  was  not 
that  we  needed  them  as  a  means  for  more  accurate  measurement, 
—  the  rational  numbers  are  entirely  adequate  for  all  mechanical 
purposes^  —  but  that  they  are  a  mathematical  necessity  if  we 
propose  to  solve  equations  of  the  type  given. 

A  similar  situation  demands  the  introduction  of  still  other 
numbers.    If  we  seek  the  solution  of 


£C2=-1, 


(1) 


we  observe  that  there  is  no  rational  number  whose  square  is  —  1. 
Neither  can  we  define  V— 1  as  a  sequence  of  rational  numbers 
which  approach  it  as  a  limit.  We  may  write  the  symbol  V— 1,  but 
its  meaning  must  be  somewhat  remote  from  that  of  V2,  for  in 
the  latter  case  we  have  a  process  by  which  we  can  extract  the 
square  root  and  get  a  number  whose  square  is  as  nearly  equal  to 
2  as  we  desire.  This  is  not  possible  in  the  case  of  V^l.  In  fact 
this  symbol  differs  from  1  or  any  real  number  not  merely  in 
degree  but  in  kind.  One  cannot  say  V— 1  is  greater  or  less  than 
a  real  number,  any  more  than  one  can  compare  the  magnitude  of 
a  quart  and  an  inch. 

V  —  1  is  symbolized  by  I  and  is  called  the  imaginary  unit.  The 
term  "imaginary"  is  perhaps  too  firmly  established  in  mathe- 
matical literature  to  warrant  its  discontinuance.  It  should  be 
kept  in  mind,  however,  that  it  is  really  no  more  and  no  less 

152 


COMPLEX  NUMBERS  153 

imaginary  than  the  negative  numbers  or  the  irrational  numbers 
are.  So  far  as  we  have  yet  gone  it  is  merely  that  which  satis- 
fies equation  (1).  When,  however,  we  have  defined  the  various 
operations  on  it  and  ascribed  to  it  the  various  characteristic 
properties  of  numbers  we  shall  be  justified  in  calling  it  a 
number. 

Just  as  we  built  up  from  the  unit  1  a  system  of  real  numbers, 
so  we  build  up  from  V—  1  =  i  a  system  of  imaginary  numbers. 
The  fact  that  we  cannot  measure  V— 1  on  a  rule  should  cause 
no  more  confusion  than  our  inability  exactly  to  measure  -y/2  on  a 
rule.  Just  as  we  were  able  to  deal  with  irrational  numbers  as 
readily  as  with  integers  when  we  had  defined  what  we  meant 
by  the  four  operations  on  them,  so  will  the  imaginaries  become 
indeed  numbers  with  which  we  can  work  when  we  have  defined 
the  corresponding  operations  on  them. 

144.  Addition  and  subtraction  of  imaginary  numbers.  We 
write 

0  =  0i, 

i  -\-  i  =  2  ij 
i  -{-  i-\ -{-  i  =  ni.  (I) 


Also  just  as  we  pass  from  a  rational  to  an  irrational  multiple 
of  unity  by  sequences,  so  we  pass  from  a  rational  to  an  irrational 
multiple  of  the  imaginary  unit.  Thus  we  write  a  V— 1,  or  ai, 
where  a  represents  any  real  number.  Consistently  with  §  76  we 
write 

^  V-a2  =  ^  Va2.(_l)  =  ±  V^ .  V^  =  ±a  V^  =  ±  ai.  (II) 

We  speak  of  a  positive  or  a  negative  imaginary  according  as 
the  radical  sign  is  preceded  by  a  positive  or  a  negative  sign. 
We  also  define  addition  and   subtraction  of   imaginaries   as 

follows : 

ai  ±  bi  =  (a  ±b)  i,  (III) 


where  a  and  b  are  any  real  numbers. 


154  ADVANCED  ALGEBRA 

Assumption.  The  commutative  and  associative  laws  of  multi- 
plication and  addition  of  real  numbers,  §  10,  we  assume  to  hold 
for  imaginary  numbers. 

145.  Multiplication  and  division  of  imaginaries.  We  have 
already  virtually  defined  the  multiplication  of  imaginaries  by 
real  numbers  by  formula  (I).    Consistently  with  §  76  we  define 

V^ .  V^  =  ii  =  i^  =  -  1. 

Thus  V—  a  •  V—  b  =  Va  •  V^  i  i  =  -\/ab  •  (—  1)  =  —  -Vab. 

The  law  of  signs  in  multiplication  may  be  expressed  verbally 
as  follows  : 

The  product  of  imar/inaries  with  like  signs  before  the  radical 
is  a  negative  real  number.  The  product  of  imaginaries  with 
unlike  signs  is  a  positive  real  number. 

For  instance,  -  yP^  -  V^^  =  -  2  •  3  •  i2  =  6. 

We  also  note  that 

i2  =  —  1,  i3  =  —  i,  i*  =  1,  i^  =r  i,  . . . , 
And,  in  general,  Hn  +  k  _  ik^  /c  =  0,  1,  2,  3. 

We  define  division  of  imaginaries  as  follows : 

/ I r  Vo^  •  i  la 

■Vb-i       ^b 

In  operating  with  imaginary  numbers,  a  number  of  the  form 
V—  a  should  always  be  written  in  the  form  Va  i  before  per- 
forming the  operation.    This  avoids  temptation  to  the  following 

error:  

V-  a  •  V-  b  =  V(-  a)  ■(-b)  =  -slab. 

EXERCISES 

Simplify  the  following : 

1.    V^8  ■  yT^. 

Solution  :   V^^  •  V^^  =  VS  •  i  .  V2  •  i  =  V2  •  8  .  {2  =  4  .  (_  1)  =  _  4. 

2.1. 

1  2^  /^  ■^—      % 

Solution :  —  =  -  = = =  —  i. 

i^      i8      (i4)2        \ 


COMPLEX  NUMBEKS 


165 


3. 

i". 

6. 

V-36. 

9. 

V-  X2«. 

12. 

V2  V-  8. 

15. 

1 

18. 

V-6 

4.    124. 

5.  ii3. 

7.   V-64. 

8.  2i'Si. 

10.   V-Sx^a^. 

11.  V-x2. 

13.   V-2V-6. 

14.  V-3% 

16.1. 

17..  f^. 

V-2 

19.  V-i2. 

20.  V-i^. 

146.  Complex  numbers.  The  solution  of  the  quadratic  equation 
with  negative  discriminant  (p.  71)  affords  us  an  expression  which 
consists  of  a  real  number  connected  with  an  imaginary  number 
by  a  H-  or  —  sign.  Such  an  expression  is  called  a  complex  number. 
It  consists  of  two  parts  which  are  of  different  kinds,  the  real 
part  and  the  imaginary  part.  Thus  6  +  4  z  means  G  I's  +  4  ^'s. 
Obviously,  to  any  pair  of  real  numbers  (x,  y)  corresponds  a  complex 
number  x  +  ty,  and  conversely. 

147.  Graphical  representation  of  complex  numbers.  We  have 
represented  all  real  numbers  on  a  single  straight  line.  When  we 
wished  to  represent  two  numbers  simultaneously,  we  made  use  of 
the  plane,  and  assumed  a  one-to-one  correspondence  between  the 
points  on  the  plane  and  the  pairs  of  numbers  (cc,  y).  The  general 
complex  number  x  -f-  iy  depends 
on  the  values  of  the  independent 
real  numbers  x  and  y,  and  may 
then  properly  be  represented  by 
a  point  on  a  plane.  We  repre- 
sent real  numbers  on  the  X  axis, 
imaginary  numbers  on  the  Y  axis, 
and  the  complex  number  x  +  iy 
by  the  point  (x,  y)  on  the  plane. 
Thus  the  complex  numbers  6  +  ^  3, 
—  4  -f  1 4,  7  —  i  5,  —  2  —  1 4  are  represented  by  points  on  the  plane 
as  indicated  in  the  figure. 

148.  Equality  of  complex  numbers.  We  define  the  two  com- 
plex numbers  a  -f  ib  and  c  -{-  id  to  be  equal  when  and  only  when 
a  =  c  and  h  —  d. 


y> 

k 

- 

4+ 

i4 

6f 

i?, 

0 

X 

-2- 

i4 

7- 

15 

rn 

156  ADVANCED  ALGEBRA 

Symbolically  a  +  ii  ^  c  +  id 

when  and  only  when  a  =  c,  b  =  d. 

The  definition  seems  reasonable,  since  1  and  i  are  different  in 
kind,  and  we  should  not  expect  any  real  multiple  of  one  to  cancel 
any  real  multiple  of  the  other. 

Similarly,  if  we  took  not  abstract  expressions  as  1  and  i  for 
units  but  concrete  objects  as  trees  and  streets,  we  should  say  that 

a  trees  +  b  streets  =  c  trees  +  d  streets 
when  and  only  when  a  =  c  and  b  =  d. 

Principle.  When  two  numerical  expressions  involving  imagi- 
naries  are  equal  to  each  other,  we  may  equate  real  parts  and 
imaginary  parts  separately. 

The  graphical  interpretation  of  the  definition  of  equality  is  that  equal  complex 
numbers  are  always  represented  by  the  same  point  on  the  plane. 

From  the  definition  given  we  see  that  a  -\-ib  =  0  when  and 
only  when  a  =  b  =  0. 

Assumption.  We  assume  that  complex  numbers  obey  the  com- 
mutative and  associative  laws  and  the  distributive  law  given  in 
§  10.  We  also  assume  the  sa.me  rules  for  parentheses  as  given 
in  §  15. 

This  assumption  enables  us  to  define  the  fundamental  opera- 
tions on  complex  numbers. 

149.  Addition  and  subtraction.  By  applying  the  assumptions 
just  made  we  obtain  the  following  symbolical  expression  for  the 
operations  of  addition  and  subtraction  of  any  two  complex  num- 
bers a  -{-ib  and  c  -{-  id: 

a  -{•  ib  ±  (c  -\-  id)  =  a  ±  c  -{-  i(b  ±  d). 

Rule.  To  add  (subtract)  complex  numbers,  add  (subtract)  the 
real  and  imaginary  parts  separately. 

150.  Graphical  representation  of  addition.  We  now  proceed 
to  give  the  graphical  interpretation  of  the  operations  of  addition 
and  subtraction. 


COMPLEX  NUMBERS 


167 


Theorem.  The  sum  of  two  numbers  A  =  a-\-  ib  and  B  =  c  -\-  id 
is  represented  hy  the  fourth  vertex  of  the  parallelogram  formed 
on  OA  and  OB  as  sides. 

Let  OASB'hQ  ?i  parallelogram.  Draw 
ES  _L  OE,  AH  A.  ES,  BD  (=  d)  _L  OE. 
A  AHS  =  A  ODE  since  their  sides  are 
parallel,  and  OB  —  AS. 

Thus 


Thus 


DB  = 

:  HS  =  d, 

0D  = 

AH==c. 

ES  = 

EH  +  HS 

Y 

1 

d         A 

7 

S 
H 

b 

0 

D 

—a, > 

JF     E   X 

b  +  d, 


OE  =  OF  +  FE  =  a  -\-  c, 

and  S  has  coordinates  (a  -\-  c,  b  -{-  d)  and  represents  the  sum  of 
A  and  B,  by  §  149. 

EXERCISES 

1.  The  difference  A  —  B  of  two  numbers  A  =  a  +  ib  and  B  =  c  +  id  is 
represented  by  the  extremity  D  of  the  line  OD  drawn  from  the  origin  par- 
allel to  the  diagonal  BA  of  the  parallelogram  formed  on  OB  and  OA  as 


2.  Represent  graphically  the  following  expressions. 

(a)  1  +  i  (b)   -  4  -  2  i       ■ 

(c)  6  -  i.  (d)  -  8  +  4  i. 

(e)  2  +  4  I  (f)  (1  +  i)  +  (2  +  i)' 

(g)  {2-i)-{6-Si).  (h)  (l_i)-(l_2i). 

(i)  (2  +  4i)-(l-3i).  (j)  4(l  +  i)-2(2-3i). 

(k)  (6-2i)  +  (2  +  3i).  (1)  (5  +  3i)  +  (-l-6i). 

151.  Multiplication  of  complex  numbers.  The  assumption  of 
§  148  enables  us  to  multiply  complex  numbers  by  the  following 

Rule.  To  multiply  the  complex  number  a  -\-  ib  by  c  ■\-  id,  pro- 
ceed as  if  they  were  real  binomials,  keeping  in  mind  the  laws  for 
multiplying  imaginaries. 

Thus  a  -\-ib 

c  -f-  id 

ac  +  icb  -f-  iad  -\-  (i)*  bd  =  ac  —  bd  +  i(cb  -f-  ad). 


158  ADVANCED  ALGEBRA 

152.  Conjugate  complex  numbers.  Complex  numbers  that  differ 
only  in  the  sign  of  their  imaginary  parts  are  called  conjugate  com- 
plex numbers,  or  conjugate  imaginaries. 

Theorem.  The  sum  and  the  product  of  conjugate  complex 
numbers  are  real  numhers. 

Thus  a  -\-  lb  -\-  a  —  ib  =  2  a, 

(a  +  ib)  (a  -  ib)  =a^  +  b\ 

153.  Division  of  complex  numbers.  The  quotient  of  two  com- 
plex numbers  may  now  be  expressed  as  a  single  complex  number. 

EuLE.    To   express    the  quotient  —  in   the  form  x  +  ^y, 

rationalize  the  denominator^  using   as  a    rationalizing  factor 
the  conjugate  of  the  denominator.  ^_^ 

™,,  a  -{-  ib       a  -\-  ib    c  —  id  

Thus  —  = ~ ^^—^ 

c  -\-  id       c  -\-  id  c  —  id 

_  ac  -\-  bd  —  i  {ad  —  be) 

~  c'  +  d^ 

ac  -{-  bd       .ad  —  be 

^  >  +  d'  ~ ""  VT^'  ^^ 

We  have  now  defined  the  fundamental  operations  on  complex 
numbers  and  shall  make  frequent  use  of  them.  If  the  question 
remains  in  one's  mind,  "After  all,  what  are  they?  "  the  answer  is 
this :  "  They  are  quantities  for  which  we  have  defined  the  funda- 
mental operations  of  numbers  and,  since  they  have  the  properties 
of  numbers,  must  be  called  numbers,  just  as  a  flower  that  has  all 
the  characteristic  properties  of  a  known  species  is  thereby  deter- 
mined to  belong  to  that  species."  Furthermore,  our  operations 
have  been  so  defined  that  if  the  imaginary  parts  of  the  complex 
numbers  vanish  and  the  numbers  become  real,  the  expression 
defining  any  operation  on  complex  numbers  reduces  to  one  defin- 
ing the  same  operation  on  the  real  part  of  the  number.  Thus  in 
(1)  above,  if  b  =  d  =  0^  the  expression  reduces  to 

a  _a 

a     c 


COMPLEX  NUMBERS  159 


EXERCISES 

Carry  out  tlie  indicated  operations. 

1.   (2  +  V"r2)(4+V^^). 
Solution  :  2  +  V^^  =  2  +  ■v/2(-l)  =  2  +  z  V2 
4  +  V:^  =:4+V5(-l)  =  4  +  t  V6 


8  -  VlO  +  i4V2  +  i2\/5 

2.  5  -  V2  -  i  Vs. 

Solution : 

5^ ^  5(V2  +  zV3)  ^  5V2  +  Z5V3  _     /g  +  i  Vs 

\/2-iV3       (V2-iV3)(V2  +  iV3)~         2  +  3 

3.  (l  +  i)l  4.  (l  +  i)3. 
Hint.  Develop  by  the  binomial  theorem. 

5.  {a  +  ib)K  6.  (V^  +  v^ir^)^ 

V7.  (x  +  %)2  8.  (x  +  i2/)2  +  (X  -  %)2. 

9.  vT+l.Vnri.  10.  (V3  +  iv^)(V2  +  iV3). 

11.  (VTT~i  +  Vr^y.  12.  (aV6  +  icVd)(aV6-icVd). 

13.  (Va  +  i  V6)  (  Va  -  i  Vb).  14.  (2  V7  +  i3  Vs)  (3  V7  -  ilOV2), 

15.  i±i^.  16.  l±i.  17.  ^ 


l-iV3  1-i  V2+V-1 

18.         ^        .  19.  11^.  20.  (^l±i^)'. 

l  +  V-3  (l  +  i)3  \         2  / 


2„    g  +  i  Vl  -  a2 
a  —  i  Vl  —  a^ 

24.  ;^-^-  25.  -^1— _.  26.  ^  +  ^p^. 

27.  ^^  +  ^^'^  28    (l±i^y.  29.  ^  +  ^^  _j_  c  +  id 


V3-iV2  \2/  a-ib      c  -id 

30.  ?? 31.  — ?i=.  32.  -V+       1 


4  +  7V^6  i  +  3v^^.  (1  +  0'    a-*y 

33    VI  +  «  +  ^  Vl-  <^  _  Vl  -  g  +  i  Vl  +  g 
Vi  -j-  g  —  i  VT  —  g      Vl  —  a  —  i  Vl  +  a 


160  ADVANCED  ALGEBRA 

34.  Find  three  roots  of  the  equation  x^  —  1  =  0  and  represent  the  roots 
as  points  on  the  plane. 

35.  Find  four  roots  of  the  equation  x*  —  1  =  0  and  represent  the  roots  as 
points  on  the  plane. 

36.  Find  six  roots  of  x^  —  1  =  0  and  represent  the  roots  as  points  on  the 
plane.    Show  graphically  that  the  sum  of  the  six  roots  is  zero. 

37.  Find  three  roots  of  x^  —  8  =  0  and  represent  the  roots  as  points  on 
the  plane.    Show  graphically  that  the  sum  of  the  three  roots  is  zero. 

154.  Polar  representation.  The  graphical  representation  of 
complex  numbers  given  in  §  147  gives  a  simple  graphical  inter- 
pretation of  the  operations  of  addition  and  subtraction,  but  the 
graphical  meaning  of  the  operations  of  multiplication  and  divi- 
sion may  be  given  more  clearly  in  another  manner.  We  have 
seen  that  we  may  represent  x  -f  iy  by  the  point  P  (x,  y)  on  the 
plane,  Represent  the  angle  between  OP  and  the  X  axis  by/^. 
This  angle  is  called  the  argument  of  the  complex  number  x  -f  iy. 

Eepresent  the  line  OP  by  p.    This  is  called 
^  ,  . ,     the  modulus  of  ic  -f  iy.    Then  from  the  figure 

a;  =  p  COS  B,  (1) 

— ^  y  =  psmO,  (2) 

x'  +  f  =  p\  (3) 

Hence  the  complex  number  x  +  iy  may  be  written  in  the  form 

X  -\-  iy  =  p  (cos  0  -}-  i  sin  0)j  (4) 

■when  the  relations  between  x,  y  and  p,  0  are  given  by  (1),  (2),  and 
(3).  A  number  expressed  in  this  way  is  in  polar  form,  and  may 
be  designated  by  (p,  6).  We  observe  that  a  complex  number 
lies  on  a  circle  whose  center  is  the  origin  and  whose  radius 
is  the  modulus  of  the  number.  The  argument  is  the  angle 
between  the  axis  of  real  numbers  and  the  line  representing  the 
modulus. 

155.  Multiplication  in  polar  form.  If  we  have  two  numbers 
p  (cos  6  -\-i  sin^)  and  /o'(cos  6'  -\-  i  sin  ^'),  we  may  multiply  them 
and  obtain 


COMPLEX  NUMBERS  161 

p (cos  6  +  isin 0)p'(Gos  0'  +  i  sin 6') 

=  pp'  [(cos  0  cos  0'  —  sin  6  sin  d') 
+  i  (sin  $  cos  d'  +  cos  0  sin  ^')] 
By  the  addition  theorem  ,r        //i   ,    /if\    ,    •    •     /h   ,    /if\n  /-i\ 

in  Trigonometry  =  PP  C^^^  (^  +  ^')  +  ^  sm  (d  +  d')]  (1) 

=  R  (cos  ©  +  *  sin  ©) .  (2) 

In  this  product  pp'  is  the  new  modulus  and  0-^0'  the  new 
argument.  We  may  now  make  the  following  statement:  The 
product  of  the  two  numbers  p  (cos  6  -\- i  sin  6)  and  p'(cos  6'  +  *  sin  0') 
has  as  its  modulus  pp'  and  as  its  argument  0  +  0'.  Thus  the 
product  of  two  numbers  is  represented  on  a  circle  whose  radius 
is  the  product  of  the  radii  of  the  circles  on  which  the  factors  are 
represented.  The  argument  of  the  product  is  the  sum  of  the 
arguments  of  the  factors. 

156.  Powers  of  numbers  in  polar  form.   When  the  two  factors 

of   the   preceding  section  (p,  0)  and  (p',  6')  are  equal,  that  is, 

when  p  =  p'  and  6  =  6',  the  expression  (1)  assumes  the  form 

[p  (cos  e  +  i  sin  0)^  =  p^ (cos  2  ^  +  t  sin  2  0).  (1) 

This  suggests  as  a  form  for  the  nth.  power  of  a  complex  number 
[p  (cos  d  -\-i  sin  $)']"  =  p«  (cos  nO  -\- i  sin  nO).  (2) 

The  student  should  establish  this  expression  by  the  method  of 
complete  induction.  The  theorem  expressed  by  (2)  is  known  as 
DeMoivre's  theorem.  Stated  verbally  it  is  as  follows  :  The  modulus 
of  the  nth.  power  of  a  number  is  the  nth  power  of  its  modulus.  The 
argument  of  the  nth  power  of  a  number  is  n  times  its  argument. 

EXERCISES 

Plot,  find  the  arguments  and  moduli  of  the  following  numbers  and  of 


eir  products. 
1.  1  +  iVS,  V3  +  i. 
Solution : 

Let            ^/S  -\-  i=p (cos ^  +  i sin 5), 
1  4-  i  V3  =  /(cos d'  +  i  sin  6^. 

2. 

30°; 

Y. 

B 

Ay    !  i     1  ! 

Then  by  (1),  (2),  (3),  §  154,  p  =  2 ;  p'  = 
1  =  2  sin  d,    hence  6  = 

0 

'--- -1— J         1      X 

1  =  2  cos  d",  hence  6'  = 

60°. 

0  =  30°,  ^'  =  60° 

162  ADVANCED  ALGEBRA 

Thus  if  the  product  has  the  form  R{cos@  +  isin©),  we  have  by  §  1( 
R  =  ppr  =  4,  ©  =  ^  +  ^'  =  90°. 


2.  l+i,2  +  i. 

3.    (l-i)3. 

4.  3  +  3t,  2^iVl2. 

5.  2i,  l-iV3. 

'4*'-^)- 

7.-1  +  1,-2-2. 

o    ,  .        V2      iV2 
^-  '^'2           2    • 

'-2+    2    '    2    +    2 

10.   [2  (cos  15°  +  i  sin  15°)]8.  11.  [i  (cos  30°  +  i  sin  30°)]4. 

12.   [|  (cos  120°  +  i  sin  120°)]2.  13.   [2  (cos  135°  +  i  sin  135°)]*. 

14.  [f (cos  180°  + isin  180°)] 3.  15.  [f (cos 315°+ isin 315°)]2. 

157.  Division  in  polar  form.  If  we  liave,  as  before,  two  com- 
plex numbers  in  polar  form  (p,  6)  and  (p',  $'),  we  may  obtain  their 
quotient  as  follows. 

p  (cos  0  -{-  i  sin  $) 
p' (cos  0'  -^  isin  0') 

_  pp'  (cos  0  -{-  i  sin  0)  (cos  0'  —  i  sin  0') 
~  p'2  (cos  ^'  +  2  sin  $')  (cos  ^'  —  i  sin  ^') 
^  pp'  [cos  (6  -  ^0  +  i  sin  (^  -  ^')] 
~  p'2(cos2^'  -f  sin^^') 


Rationalizing, 


§  152  and  §  153, 


Since  sin2  0  +  cos2  6=1,  =  -^  [cos  (0  —  $')  +  i  sin  (^  —  0')'] 

=  i2  (cos^Vf  i  sin  ©). 

We  may  now  make  the  following  statement:  The, quotient  of 
two  complex  numbers  has  as  its  modulus  the  quotient  of  the  moduli 
of  the  factors,  and  as  its  argument  the  difference  of  the  arguments 
of  t/ie  factors. 

158.  Roots  of  complex  numbers.  We  have  seen  that  the  square 
of  a  number  has  as  jts  modulus  the  square  of  the  original  modulus, 
while'  the  argument  is  twice  the  original  argument. 

Thjs  would  suggest  th^-t  the  square  root  of  any  number,  as  (p,  0), 

'  :        '         -  9 

would  have  Vp  as  its  modulus  and  -  as  its  argument.    Since 

'^  ^ 

every  yeal  number  has  two  square  roots,  we  should  expect  the 

same  fact  to  hold  liere.    Consider  the  two  numbers 


COMPLEX  NUMBERS 


163 


Vp  /^cos  I  +  i  sin  I  j  and  Vp  cos  (^  -f  180°  j  +  i  sin  ( |  +  180°  j  j  , 

where  Vp  is  the  principal  square  root  of  p(§  72).  The  square  of 
the  first  is  (p,  ^),  by  §  155.  That  the  square  of  the  second  is  the 
same  is  evident  if  we  keep  in  mind  the  fact  that 

cos  (0  +  360°)  =  cos  e 
and  sin  (^  +  360°)  =  sin  a 

Thus  V/>  (cos  e  +  i  sin  6) 


Vp  (  cos  -  +  i  SlUv- 


or 


Vp 


+  180°    -f  j  sin/ ^  +  180 


■)]■ 


The  graphs  of  these  two  numbers  are  situated  at  points  sym- 
metrical to  each  other  with  respect  to  the  origin. 

We  may  obtain  as  the  corresponding  expression  for  the  higher 
roots  of  complex  numbers  the  following : 


■v^/>(cos^  +  ?:sin^)=  V^  I  ( 


(9  +  /c360°\      .   .    /^  +  A;360' 

cos  I +  I  sin 

j_       \         71         /  \         n 


where  for  a  given  value  of  /i,  k  takes  on  the  values  0,  1,  •  •  •,  t^  —  1, 
and  where  VP  indicates  the  real  positive  7ith  root  of  p. 


EXERCISES 

Perform  the  indicated  operations  and  plot 
1.  2-2V3i--l+i. 
Solution : 

Let  \-\-i  —  p  (cos  0  +  i  sin  ^), 

2  --  2  V3  i  =  /(cos  e'-\-  i  sin  6"). 
Then  p  =  \/12  +  1-  =  V2, 

/  =  V22  +  (-2V3)^  =  4. 


By  (1)  and  (2),  §  154, 

sin  ^  =  cos  ^  =  — ^ ,  hence  6  =  46°. 
V2 


'4; 


'^-^ 


(a-iaW; 


164 

Similarly, 


ADVANCED  ALGEBliA 


300° 


sin^  = 

- 

2V3 
4 

V3 
2 

cos^  = 

f 

=  1,  hence  d'  - 

2  _  2  V3  i  4  (cos  300°  +  i  sin  300°) 

Thus        =  i2(cos©  +  isin©)=— 7^- -— .  .    ,^^  ■ 

1+i  ^  '      V2(cos46°4- tsin45°) 

Hence  by  §  157,    iJ  =  -^  =  2v^,  ©  =  300°  -  45°  =  255» 

V2 


/ 


2.  V-2  +  2V3i 
Let      -2  +  2  V3i  =  p(cos^4'isin^). 
Then  (§  154)  />  =  4,  cos^=  -  f  =  -^, 


-2+i2A^ 


and 


120^= 


V-2  +  2V3i  =  V4  (cos  120°  +  i  sin  120°) 

T5    «i^Q                      /if       /120°  +  A:360°\ 
By  §158,  =V4|cos( ^ j 


l+iVs" 


.  .    /120°  +  A;360°\'] 
.sm( ^)J 


+  ^ 
(where  A;  =  0  or  1) 

=  2  (cos 60°  +  i  sin 60°)  =  1  +  i  V3,  when  k-0. 
=  2  (cos  240° + i  sin  240°)  =  - 1  -  i  V3,  when  A; = I 

.     3.   VV2  +  i  V2. 


4.  V 


1  +  i  -^  1  -  i. 
1       V3 


6. 


1  iV3 

2  2 


^1  +  i. 


8    -l-i-^-^l  +  l  N 

2         2      ■  4      4* 

10.   -2\/2-2  V2i-T--2  +  2V3i 


7.  i  +  i^:t_j:Jl 
2        2 

9.  2-iVl2-4-3  +  3i. 

11.   \^l(cosl5°+isinl5°). 
Solution : 

"/TT ^:,o   .     -     -     -.  ^Q^  '/tF         /15°  +  fe  •  360°\    .     .     .      /16°+ fc  •  360°\  1 

Vl(cosl5°+  ism  15°)  =  VI    cosi ■ )  +  ism  I 1 

(where  fc  =  0,  1,  or  2). 

1  (cos  5°  +  t  sin  5°),  when  A;  =  0, 
1  (cos  125°  +  i  sin  125°),  when  A;  =  1, 
1  (cos  245°  +  i  sin  245°),  when  A;  =  2. 

12.  Yi. 

13.  ^^161. 

14.  •V'2-f- 2\/3i. 

15.  v'cos330°+  tsin330<». 


COMPLEX  KUMBERS 


165 


16.   V27(cos75°+ism75°).  17.   Vl6  (cos  200°  +  i  sin  200°). 

18.  Solve  the  following  equations  and  plot  their  roots. 

(a)  x5  -  1  =  0. 

Solution :  x^  =  1,  or  a;  =  VT. 

Let  1  =  1  +  0  •  i  =  p  (cos  ^  +  i  sin  6).    Then  p  =  l,  6  =  0°. 


X  =  V 1  (cos  0°  4-  i  sin 


0°)  =  Vircos/* 


0°  +  A; .  360' 


(where  k  takes  on  the  values  0,  1,  2,  3,  4) 

f  cos     0°  +  i  sin  0°  =  1,  when  k  =  0, 
cos   72°  +  i  sin  72°,  when  A;  =  1, 
cos  144°  +  i  sin  144°,  when  fe  =  2,  • 
cos  216°  +  t  sin  216°,  when  A;  =  8, 
cos  288°  +  i  sin  288°,  when  A;  =  4. 

These  numbers  we  observe  lie  on  a  circle  of 
unit  radius  at  the  vertices  of  a  regular  pentagon. 


(b)  a;4  -  1  =  0. 
(e)  x6  -  1  =  0. 

J 


(C)    X3-1  =  0. 
(f )   X8  -  1  =  0. 


C?x~^  <Jtf^.C<. 


(d)  x5  -  32  =  0. 
(g)  x8  --  27  =  0. 


J^l^^ 


y. 


O^ 


CHAPTEE  XVII 
THEORY  OF  EQUATIONS 

159.  Equation  of  the  nth  degree.  Any  equation  in  one  variable 
in  whicli  the  coefficients  are  rational  numbers  can  be  put  in  the 

form 

f (x)  =aoX'^  +  a^x--' -}--'•  + a,  =  0,  (1) 

where  Gq  is  positive  and  a^,  •  •  •,  a„  are  all  integers. 

The  symbol /(x)  is  read  "foix"  and  is  merely  an  abbreviation  for  the  right- 
hand  member  of  the  equation.  Often  we  wish  to  replace  x  in  the  equation  by 
some  constant,  as  a,  —  2,  or  0.  We  may  symbolize  the  result  of  this  substitution 
by/(a),/(-2),or/(0). 

Thus  f{b)  =  a(fi»  +  aift""  ^  +  ^  •  •  +  a„ . 

We  symbolize  other  expressions  similarly  by  <f)  (x),  Q{x),  etc. 

When  we  speak  of  an  equation  we  assume  that  it  is  in  the  form 
of  (1).    This  equation  is  also  written  in  the  form 

a;"  +  M"-i  +  ---^„  =  0,  (2) 

where  bi  =  —>  b^  =  —>  ••  •  K  —  ~' 

The  5's  are  integers  only  when  ai,  «2j  •  •  -,  «„  are  multiples  of  Qq. 

160.  Remainder  theorem.  We  now  prove  the  following  impor- 
tant fact. 

Theorem.  When  f(x)  is  divided  hy  x  —  c,  the  remainder  is 
f(x)  with  c  substituted  in  place  of  the  variable. 

Divide  the  equation  (1)  by  x  —  c.  Let  R  be  the  remainder, 
which  must  (§  26)  be  of  lower  degree  in  x  than  the  divisor ; 
that  is,  in  this  case,  since  x  —  c  is  the  divisor,  R  must  be  a  con- 
stant and  not  involve  x  at  all.  Let  the  quotient,  which  is  of 
degree  n  ~  1  in  x,  he  represented  by  Q  (x). 

160 


THEORY  OF  EQUATIONS  167 

Then  i^  =  Q(x)+  -5— 

X  —  C  ^     ^        X  —  G 

Clearing  of  fractions, 

f(x)=  Q(x){x-g)+R. 

But  since  this  equation  is  an  identity  it  is  ahvays  satisfied 
whatever  numerical  value  x  may  have  (§  53). 

Let  X  =  c. 

Then  /(c)  =  a,c-  +  %c"-i  +  ...  +  «„=  Q  (c)  (c  -  c)  +  i2. 

But  since  c  —  c  =  0,  Q(c)  (c  —  c)  =  0,  and 

R  =  aoc"  +  ^ic"- 1  H h  a„  =/(c)- 

Corollary.  If  c  is  a  root  of  f(x)  =  0,  then  x  —  c  is  a  factor 
of  the  left-hand  member.  , 

For  if  c  is  a  root  of  the  left-hand  member,  it  satisfies  that 
member  and  reduces  it  to  zero  when  substituted  for  x.  Thus  by 
the  previous  theorem  we  have,  since 

aoC*  H-  o^ic"-^  H a^  =  R  =  0, 

f{x)=Q(x)(x-c). 

161.  Synthetic  division.  In  order  to  plot  by  the  method  of 
§  103  the  equation 

y  =  a^x""  +  a^x""- 1  H \-  a^, 

when  the  a's  are  replaced  by  integers,  we  should  be  obliged 
laboriously  to  substitute  for  x  successive  integers  and  find  corre- 
sponding values  of  y,  which  for  large  values  of  n  involves  con- 
siderable computation.  We  can  make  use  of  the  preceding  theorem 
to  lighten  this  labor.  The  object  is  to  find,  with  the  least  possible 
computation,  the  remainder  when  the  polynomial  f(x)  is  divided 
by  a  factor  of  form  x  —  c,  which  by  the  preceding  theorem  is  the 
value  of  f(x)  when  x  is  replaced  by  c,  that  is,  the  value  of  y 
corresponding  to  x  =  c.    For  illustration,  let 

f(x)  =  2  x^  -  3  a;«  -h  ?c2  -  a:  -  9  and  c  =  2. 


168  ADVAIJCED  ALGEBKA 

By  long  division  we  have 

a;-2|2x^-3a;^H-     ^' -     x-    9|2a;«  +  o^' +  3a;  +  5 

1  x»  +     a;2 
lx3-2a;2 


3a;2-     X 

3a;2-6x 


5a;-    9 
5a; -10 

+    1 

We  can  abbreviate  this  process  by  observing  the  following 
facts.  Since  x  is  here  only  the  carrier  of  the  coefficient,  we  may 
omit  writing  it.  Also  we  need  not  rewrite  the  first  number  of 
the  partial  product,  as  it  is  only  a  repetition  of  the  number 
directly  above  it  in  full-faced  type.  Our  process  now  assumes 
the  form 

1- 212-3  +  1-1-    9|2  +  l  +  3  +  5 

:l-4 
+ 1  -?  / 

/  -2 


+  3       ,' 
-6 


+  5 


10 


+    1 

Since  the  minus  sign  of  the  2  changes  every  sign  in  the  partial 
product,  if  we  replace  —  2  by  +  2  we  may  add  the  partial  prod- 
uct to  the  number  in  the  dividend  instead  of  subtracting.  This 
is  also  desirable  since  the  number  which  we  are  substituting  for 
X  is  2,  not  —  2.  Thus,  bringing  all  our  figures  on  one  line  and 
placing  the  number  substituted  for  x  at  the  right  hand,  we  have 

2-3  +  1-1-    9[2 

+  4  +  2  +  6  +  10 
2+1+3+5+    1 


THEORY  OF  EQUATIONS  169 

We  observe  that  the  figures  in  the  lower  line,  2,  1,  3,  5,  up  to 
the  remainder  are  the  coefficients  of  the  quotient  2  x^  -{- x^  -\- S  x -\- 5. 

EuLE  FOR  SYNTHETIC  DIVISION.  Write  the  coefficients  of  the 
polynomial  in  order,  supplying  0  when  a  coefficient  is  lacking. 

Multiply  the  number  to  he  substituted  for  x  by  the  first  coeffi- 
cient, and  add  (algebraically)  the  product  to  the  next  coefficient 

Multiply  this  sum  by  the  number  to  be  substituted  for  x,  add  to 
the  next  coefficient,  and  proceed  until  all  the  coefficients  are  used. 
The  last  sum  obtained  is  the  remainder  and  also  the  value  of 
the  polynomial  when  the  number  is  substituted  for  the  variable. 

162.  Proof  of  the  rule  for  synthetic  division.  This  rule  we  now 
prove  in  general  by  complete  induction.    Let  the  polynomial  be 

a^x""  4-  a^x^-^  +  a^x""-^  -\ h  «„. 

Let  the  number  to  be  substituted  for  x  be  a. 

First.   Let  n  =  2.    Carry  out  the  rule  on  a^x"^  +  a^x  +  a^. 

We  have  ,  ,  . 

+  aQ(X         +  {apa  +  a\)  cc 

<^oi  +  «o«^  +  «i,  +  {a^a  -|-  a.i)  a  +  ^2  =  a^a^  +  a^a,  +  a^. 

Second.   Assume  the  validity  of  the  rule  for  n  =  m,  and  prove 

that  its  validity  f or  ti  =  m  +  1  follows.    Assume  then  that  the  rule 

carried  out  on 

f{x)  =  a^x^  +  a^x^-^  +  •••  +  «„ 

affords  the  remainder 

a^a"^  +  ai^"*-!  ^ -\- a^  =f((x)' 

Now  the  polynomial  of  order  w  +  1  is 

^0^"'+^  4-  aix""  H \-a^x-i-  a^  +  j  =  x  •f(x)  +  a^^^ 

Hence  the  next  to  the  last  remainder  obtained  by  applying  the 
rule  to  this  polynomial  would  be  f(cc),  since  the  succession  of 
coefficients  is  the  same  for  both  polynomials  up  to  a^+i-  By 
the  rule  the  final  remainder  is  obtained  by  multiplying  the  expres- 
sion just  obtained,  in  this  case  f(cc),  by  a  and  adding  the  last 
coefficient,  in  this  case  a^_^,^.    This  affords  the  final  remainder 


170 


ADVANCED  ALGEBRA 


EXERCISES 

1.  Prove  by  complete  induction  that  the  partial  remainders  up  to  the  final 
remainder  obtained  in  the  process  of  synthetic  division  are  the  coefficients  of 
the  quotient  of  f{x)  hj  z  —  a. 

2.  Perform  by  synthetic  division  the  following  divisions. 

(a)  a;3  -  7x2  _  6x  +  72  by  a;  -  4. 

Solution :  1-7-6  + 72  [4 

4  _  12  -  72 


1  _  8  -  18        0 
Quotient  =  a;2_3x- 18. 

(b)  aj8  -  9x  +  10  by  X  -  2.  (c)  4x3  -  7 x  -  87  by  x  -  3. 

(d)  x3  +  8x2  -  4x  -  32  i3y  x-2.      (e)  x^  +  4x2  -  7x  -  30  by  x  +  3. 

(f)  x8  -  6x2  +  11 X  -  6  by  X  -  1. 

(g)  x*  -  16x3  +  86x2  -  176x  +  105  by  x2  -  8x  +  7. 

Hint.    SinceK2— 8a;+7=(a;  — 7)(x  — 1),  divide  by  x—7  and  the  quotient  by  a;— 1. 

(h)  x6  +  1  by  X  +  1.  (i)  x9  -  1  by  X  -  1. 

(j)  x4  +  x3  -  X  -  1  by  x2  -  1.  (k)  x^  -  2x3  -  4x  by  x  -  3. 

(1)  x6  -  2x8  -  4x  -  1  by  X  +  2.        (m)  4x3  -  6x2  -  2x  -  1  by  x  -  3. 

(n)  2x*  +  5x3  -  37x2  +44x  +  84  by  x2  +  6x  -  6. 

163.  Plotting  of  equations.    We  can  now  form  the  table  of 
values  necessary  to  plot  an  equation  of  the  type 

^0^"  +  ^i^'""  ^  H 1-  «„-i^  +  «„  =  y. 

Example.    Plot  x3  +  4  x2  -  4  =  j/. 
l  +  4  +  0-4[l 

+1+5+5 
1+6+6+1 
1  +  4  +  0-41-1 

-1-3+3 
l+S-S-l 

1  +  4  +  0-41-2 

-2-4+8 
1+2-4+4 
l_f.4  +  0-4[-_3 

-3-3+9 

+1-3+5 
1  +  4  +  0-41-4 

-4+0+0 


X 

V 

X 

y 

0 

-4 

-  1 

-1 

1 

+  1 

-2 

+  4 

;l  xo 

-3 

+  6 

-4 

-4 

1+0+0-4 


THEORY  OF  EQUATIONS  171 

In  this  figure  two  squares  are  taken  to  represent  one  unit  of  x.  A  single 
square  represents  a  unit  of  y. 

By  an  inspection  of  the  figure  it  appears  that  the  curve  crosses  the  X  axis 
at  about  x  =  .8,  x  =  —  1.2,  and  x  =  —  3.7.  Thus  the  equation  for  y  =  0  has 
approximately  these  values  for  roots  (§  110). 

164.  Extent  of  the  table  of  values.  Since  the  object  of  plot- 
ting a  curve  is  to  obtain  information  regarding  the  roots  of  its 
equation,  stretches  of  the  curve  beyond  all  crossings  of  the  X  axis 
are  of  no  interest  for  the  present  purpose.  Hence  it  is  desirable 
to  know  when  a  table  of  values  has  been  formed  extensive  enough 
to  afford  a  plot  which  includes  all  the  real  roots.  If  for  all  values 
of  X  greater  than  a  certain  number  the  curve  lies  wholly  above 
the  axis,  there  are  no  real  roots  greater  than  that  value  of  x. 

By  inspection  of  the  preceding  example  it  appears  that  if  for 
a  given  value  of  x  the  signs  of  the  partial  remainders  are  all 
positive,  thus  affording  a  positive  value  of  ?/,  any  greater  value 
of  X  will  afford  only  positive  partial  remainders  and  hence  only 
positive  values  of  y. 

Thus  when  ■  all  the  partial  rem.ainders  are  positive  no  greater 
positive  value  of  x  need  he  substituted. 

Similarly,  when  the  partial  remainders  alternate  in  sign  begin- 
ning with  the  coefficient  of  the  highest  power  of  x,  no  value  of  x, 
greater  negatively,  need  be  substituted. 

In  plotting,  if  the  table  of  values  consists  of  values  that  are 
large  or  are  so  distributed  that  the  plot  would  not  be  well  propor- 
tioned if  one  space  on  the  paper  were  taken  for  each  unit,  a  scale 
should  be  so  chosen  that  the  plot  will  be  of  good  proportion, 
that  is,  so  that  all  the  portions  of  the  curve  between  the  extreme 
roots  shall  appear  on  the  paper,  and  the  curvatures  shall  not  be 
too  abrupt  to  form  a  graceful  curve.  This  was  done,  for  example, 
in  the  figure,  §  163. 

EXERCISES 

Plot  and  measure  the  values  of  the  real  roots  of  the  equations  when  y  =  0. 
1.  x8  -  7  X  -  6  =  y.  2.  x3  -  7  X  +  5  =  y. 

3.  7x8  -  9x  -  6  =  y.  4.  x3  -  31 X  +  19  =  y. 

6.  x3  -  12x  -  14  =  y.  6.  4x8  -  13x  +  6  =  y. 


172  abvakced  algi:biia 

7.  a;8  -  12x  -  16  =  y.  8.  x^  -  45x  +  152  =  y. 

9.  x*  -  2x3  -  X  +  2  =  y.  10.  8x3  _  igxS  +  17x  -  6  =  y. 

11.  X*  -  17x2  +  X  +  20  =  y.  12.  x-i  -  4x3  +  9x2  -  8x  +  14  =  y. 

13.  18x3-36x2  + 9x  + 8  =  2/.  14.  x*  +  5x3  +  12x2  + 52x- 40  =y. 

15.  x*-2x3-7x2+19x-10=y.       16.  x4-6x3  +  3x2  +  26x-24  =  y. 
17.  6x4  -  13x3  +  20x2  -  37x  +  24  =  y. 

165.  Roots  of  an  equation.  In  the  case  of  the  linear  and 
quadratic  equations  we  have  been  able  to  find  an  explicit  value 
of  the  roots  in  terms  of  the  coefficients.  Such  processes  are  prac- 
tically impossible  in  the  case  of  most  equations  of  higher  degree. 
In  fact  the  proof  that  any  equation  possesses  a  root  lies  beyond 
the  scope  of  this  book,  and  we  make  the 

Assumption.   Every  equation  possesses  at  least  one  root. 

This  is  equivalent  to  the  assumption  that  there  is  a  number, 
rational,  irrational,  or  complex,  which  satisfies  any  equation. 

166.  Number  of  roots.  We  determine  the  exact  number  of 
roots  by  the  following 

Theorem.    Every  equation  of  degree  n  has  n  roots. 

Given  the  equation /(x)  =  ao^?"  +  aiic""^  -\ +-  a„  =  0. 

Let  ^i  (see  assumption)  be  a  root  of  this  equation.  Then  (p.  166) 
ic  —  ^i  is  a  factor  of  the  left-hand  member,  and  the  quotient  of 
f{x)  by  £c  —  «!  is  a  polynomial  of  degree  n  —1.    Suppose  that 

aoic"  -f-  «ia;«- 1  -f  •••  +  ««  =  «o(^  -  «^i)  (ic""  ^  +  ^i^c""^  H h  ^„- 1). 

By  our  assumption  the  quotient  a;""  ^  -f  Jicc'*"^  4-  •  •  •  +  ^„_  i  =  0 
has  at  least  one  robt,  say  az,  to  which  corresponds  the  factor 
X  —  a^.    Thus 

f{x)  =aQ{x-  a^  {x  —  a^)  (a;"-^  -\-  CiX"-^  -\ h  c^.g). 

Proceeding  in  this  way  we  find  successive  roots  and  corre- 
sponding linear  factors  until  the  polynomial  is  expressed  as  the 
product  of  n  linear  factors  as  follows : 

f{x)  =ao(x-  ai)  (x-a2)'"(x-  a„)  =  0, 
where  the  roots  are  Uu  ag,  •  •  • ,  a„. 

Remark.  This  theorem  gives  no  information  regarding  how  many  of  the  roots 
may  be  real  or  imaginary.   This  depends  on  the  particular  values  of  the  coefl&cients. 


THEORY  OF  EQUATIONS  173 

Corollary.  Any  polynomial  in  x  of  degree  n  may  he  expressed 
as  the  product  of  n  linear  factors  of  the  form  x  —  a,  where  a  is 
a  real  or  a  complex  number. 

It  should  be  noted  that  the  roots  are  not  necessarily  distinct. 
Several  of  the  roots  and  hence  several  of  the  factors  may  be 
identical. 

If  f(x)  is  divisible  by  {x  —  aiY,  that  is,  if  a^  =  a^,  we  say  that 
oTi  is  a  double  root  of  the  equation.  Similarly,  \i  f(x)  is  divis- 
ible by  {x  —  oTi)'',  a^  is  called  a  multiple  root  of  order  r.  When 
we  say  an  equation  has  n  roots  we  include  each  multiple  root 
counted  a  number  of  times  equal  to  its  order. 

Theorem.  An  equation  of  degree  n  has  no  more  than  n 
distinct  roots. 

Let/(x)  =  ^0^"  H f"  ^n  =  ^  have  the  roots  a-^^,  cc2j'"j  *».•   Write 

the  equation  in  the  form 

ao(x  —  ai)  •  • '  (x  —  a^)  =  0. 

If  r  is  a  root  distinct  from  a^,  •••,«:„,  it  must  satisfy  the  equation 
and 

ao(r-aj)"'(r-a^)  =  0. 

Since  this  numerical  expression  vanishes  one  of  its  factors 
must  vanish  (§5).  But  r  =f=  ai,  thus  r  —  a^  =^  0.  Similarly,  no 
one  of  the  binomial  factors  vanishes.  Thus  (§5)  «<,  —  0,  which 
contradicts  the  hypothesis  that  the  equation  is  of  degree  n. 

This  theorem  may  also  be  stated  as  follows : 

Corollary  I.    If  an  equation  a^x"  +  a^^x""  "^  + \-  a^=  0  of 

degree  n  is  satisfied  by  more  than  n  values  of  x,  all  its  coefficients 
vanish. 

The  proof  of  the  theorem  shows  that  if  the  equation  has  n  +  1 
roots,  ao  =  0.  We  should  then  have  remaining  an  equation  of 
degree  n —1,  also  satisfied  hj  n  +1  values  of  x.  Thus  the  coeffi- 
cient of  its  highest  power  in  x  vanishes.  Similarly,  each  of  the 
coefficients  vanishes. 


174  '  ADVANCED  ALGEBRA 

Corollary  II.  If  two  ^polynomials  in  one  variable  are  equal 
to  each  other  for  every  value  of  the  variable^  the  coefficients  of 
like  ^powers  of  the'  variable  are  equal  and  conversely. 

Let  a^x""  +  a^x""-^  H h  o^„  =  ^o^"  +  ^i^c""^  H h  *„ 

for  every  value  of  x. 

Transpose,       {a^  —  h^x''-\--'--\-a^  —  h^  =  ^. 

By  Corollary  I,  a^  —  h^^^  0,  or  a^  =  b^, 

ai  —  bi  =  0,  or  a^  =  bi, 

(^n-K  =  0,  or  a„  =  ^>„. 

167.  Graphical  interpretation.  The  graphical  interpretation 
of  the  theorems  of  the  preceding  section  is  that  the  graph  of  an 
equation  of  degree  n  cannot  cross  the  X  axis  more  than  n  times. 
Since  each  crossing  of  the  X  axis  corresponds  to  a  real  root,  there 
will  be  less  than  n  crossings  if  the  equation  has  imaginary  roots. 

168.  Imaginary  roots.  We  now  show  that  imaginary  roots 
occur  in  pairs.    This  we  prove  in  the  following 

Theorem.  If  a  -\-  ib  is  a  root  of  an  equation  with  real  coeffi- 
cients, a  —  ib  is  also  a  root  of  the  equation. 

li  a  -{-  ib  is  a  root  of  the  equation  <Xo^"  +  a^x^'  ^  -\-  -••-{-  a^  =  0^ 
then  X  —(a  -{-  ib)  is  a  factor  (p.  166).  We  wish  to  prove  that 
X  —  {a  —  ib)  is  also  a  factor,  or  what  amounts  to  the  same  thing, 
that  their  product 

[x  —{a-\-  ib)'\  [x  —{a  —  lb)']  =  [(x  —  a)  —  ib"]  [(x  —  a)-\-  ib"] 

is  a  factor  of  f(x).    Divide  f(x)  by  (x  —  a)^  +  b^  and  we  get 

f(x)  =  Q(x)  [(x  -  af  J^b^-]J^rx  +  r',  (1) 

where  r  and  r'  are  real  numbers.    This  remainder  rx  +  r'  can  be 
of  no  higher  degree  in  x  than  the  first,  since  the  divisor 

(x  -  af  +  b'' 


THEORY  OF  EQUATIONS 


175 


is  only  of  the  second  degree  (§  26).  Now  this  equation  (1)  being 
an  identity  is  true  whatever  value  is  substituted  for  x,  as,  for 
instance,  the  root  of  /(x),  a  -\-  ib.  Substituting  this  value  for  x, 
we  get  i 

f{a  Jrib)  =  0  =  Q{a  +  ib)  \_{a  +  ib  -  ay  +  b^-]^^  r  {a -\-  ib)  +  r', 

or  (p.  33)  and  (p.  3)  0  =  0  +  r«^  -f  r'  +  irb, 

or  (p.  156)  m  +  r'  =  0,  (2) 

rb  =  0.  (3) 

Since  b  ^  0,  by  (3),  p.  3,  r  =  0. 

Also  from  (2),  r'  =  0.  > 

Hence  rx  -{-  r'  =  0. 

Consequently  there  is  no  remainder  to  the  division  of  f(x)  by 
(x  —  a)^  -\-  U\  and  hence  if  a  +  ib  is  a  root  oif(x),  a  —  ib  is  also  a 
root. 

Corollary.  Every  equation  of  odd  degree  with  real  coeffi- 
cients has  at  least  one  real  root. 

The  roots  cannot  all  be  imaginary,  else  the  degree  of  the  equa- 
tion would  be  even  by  the  preceding  theorem. 

169.  Graphical  interpretation  of  imaginary  roots.  When  we 
plot  the  equations 


y  =  x^  +  Ax''-4.  (1), 
Yk 


y  =  x^-i-4:x^-l  (2), 


176 


ADVANCED  ALGEBRA 


7/ =  ^^  +  4^2  (3), 


y 

= 

a;8 

+  4 

x" 

+1  (4 

0. 

^ 

\ 

/ 

/ 

\ 

/ 

/ 

>v 

/ 

r 

\ 

/ 

^ 

\ 

\ 

\ 

\ 

\ 

\ 

J 

x^ 

, 









0 

— 

— 

X 

we  see  that  corresponding  to  the  increase  of  the  constant  term  is 
a  corresponding  elevation"  of  the  curve  with  respect  to  the  X  axis. 
In  fact  in  each  case  the  curve  is  the  same,  but  the  value  of  y  is 
gradually  increased.  In  (1)  and  (2)  we  have  three  real  roots,  in 
(3)  the  curve  touches  the  X  axis,  and  in  (4)  we  have  only  one 
real  root.  As  the  elbow  of  the  curve  is  raised  and  fails  to  intersect 
the  X  axis  a  pair  of  roots  cease  to  be  real,  and  since  a  cubic  equa- 
tion always  has  three  roots,  a  pair  of  roots  become  imaginary. 
Thus  we  have  the 

Pkinciple.  Corresponding  to  every  elbow  of  the  curve  that 
does  not  intersect  the  X  axis  there  is  a  pair  of  imaginary  roots 
of  the  equation. 

The  converse  is  not  always  true.  It  is  not  always  possible  to 
find  as  many  elbows  of  the  curve  which  do  not  meet  the  X  axis 
as  there  are  pairs  of  imaginary  roots. 


EXERCISES 

Plot  the  following  equations  and  determine  from  the  plot  how  many  roots 
are  real. 

1.  X*  -  1  =  y.  2.  a;6  -  2  =  y.  3.  x^  -  x  -  1  =  y. 

4.  x4  +  l  =  y.  5.  x*  +  x  +  l  =  ?/.  6.  x*  +  2x2  +  2  =  y. 

7.  x8  -  3x2  -  X  +  1  =  y.  8.  x8  -  2x2  +  4x  -  1  =  y. 

9.  2x8  +  3x2  +  6x  +  6  =  y.  10.  x^  -  3x2  -  4x  -  5  =  y. 


THEORY  OF  EQUATIONS  177 

170.  Relation  between  roots  and  coefficients.  If  we  write  the 
expression  (Corollary,  p.  173) 

and  multiply  the  factors,  we  obtain  by  equating  coefficients  of  like 
powers  of  x  (p.  174)  relations  between  the  roots  and  the  coeffi- 
cients.   Take  for  example  n  =  S. 

x'  +  Wx^  +  hx  +  b,  =  (x-  13,)  (x  -  13,)  (X  -  p,) 

=  x'-  (13,  -{-/3,  +  13,)  x'  +  ()8i/82  +  ^2^3  +  ft  A)  X  -  )8i  A^3  =  0. 
.    Hence  j^  =  _  (/3^  +  ^^  +  ^3), 

*2  =  Aft  +  Aft  +  A  A, 

^3  =  -AAft- 

This  suggests  the 

Theorem.  The  coefficient  of  ^""^  is  equal  to  the  sum  of  the 
roots  with  their  signs  changed. 

The  constant  term  is  equal  to  the  product  of  the  roots  with 
their  signs  changed. 

In  general  the  coefficient  of  a?""*"  is  equal  to  the  sum  of  all 
possible  products  of  r  of  the  roots  with  their  signs  changed. 

We  prove  this  theorem  by  complete  induction. 
First.   We  have  already  established  the  theorem  for  equations 
of  degree  two  on  p.  106  and  for  equations  of  degree  three  above. 
Second.   Assume  the  theorem  for  n  =^m.    That  is,  if 

£c-  +  ^'1^'"-'  +  •  •  •  +  *«.  =  (^  -  A)  (^  -  A)  •  •  •  (^  -  A).  (1) 
we  assume  that  h^,  the  coefficient  of  ic'"-'",  is  the  sum  of  all  possi- 
ble products  of  r  of  the  numbers  —  A?  —  A?  '  • '?  ~  A- 

Multiply  both  sides  of  (1)  hy  x  —  A«  +  i.  Denote  the  result  by 
^m+i  +  b^^x^  +  . . .  +  5'^^^  =  (x-  p,)(x  ^p,)...(x-  ^^  +  0-   (2) 

The  term  in  0;"'+^-'"  in  this  equation  is  obtained  by  multiplying 
the  terms  b.x'^-'-  and  &^_ia;'»+^-''  in  (1)  by  x  and  -  p^+i  respec- 
tively.   That  is,  in  (2) 

b'r  =  K  +  b^-,(-(im^O'  (3) 

Now  all  possible  products  of  r  of  the  quantities  —  A>  —ft* 
..  •,  —  A 4.1  may  be  formed  as  follows  :  (1)  Neglect  —  ft  +  i,  and 
form  all  possible  products  of  r  of  those  remaining.    The  sum  of 


178  ADVANCED  ALGEBRA 

these  is  b^.    (2)  Form    all  possible   products  of  r  -  1  of   —  )8i 
—  Aj  •  •  •>  —  Pm^  not  including  —  Pm  +  u  and  multiply  each  product 
^y  —  Pm+\-    Add  all  the  products  obtained.    This  process,  it  is 
observed,  is  precisely  that  indicated  by  (3). 

Remark.  It  is  noticed  that  in  the  rule  the  signs  of  the  roots  are  always  changed 
before  forming  any  term.  This  does  not  involve  any  change  when  r  is  an  even 
number,  but  is  included  in  the  rule  for  the  sake  of  uniformity. 

Corollary.  Every  root  of  an  equation  is  a  factor  of  its  con- 
stant term. 

171.  The  general  term  in  the  binomial  expansion.   On  p.  129 

we  gave  an  expression  for  the  (r  +  l)st  term  of  the  binomial  ex- 
pansion, the  validity  of  which  we  now  establish.  In  (1),  §  170, 
let  ySi  =  ^2  =  •  •  •  =  Pn-  Denote  this  common  value  by  —  a.  The 
expression  (1)  becomes,  on  writing  n  in  place  of  m, 

X-  +  hx""-'  +  ---  +  b,=(x  +  ay. 
By  the  theorem  in  §  170,  b^  is  the  sum  of  all  possible  products 
of  r  of  the  negative  roots.    Since  there  are 


a.= 


n(n  —  1)  "•  (n  —  r  -i-l) 


ri 


such  products,  and  since  the  roots  are  now  identical,  we  obtain 
n(n-l)>-'(n-r-{-l)  ^^_^^^ 
rl 
as  the  form  of  the  (r  +  l)st  term  of  the  expansion  of  (x  +  a)". 

172.  Solution  by  trial.  Since  by  the  previous  corollary  every 
root  of  an  equation  is  a  factor  of  its  constant  term,  we  may  in 
many  cases  test  by  synthetic  division  whether  or  not  a  given  equa- 
tion has  integral  roots.    Thus  the  integral  roots  of  the  equation 

ic*  _  8a;»  +  4:ic=»  -f  24a;  -  21  =  0  (1) 

must  be  factors  of  21. 

We  try  +  1  by  synthetic  division,  ' 

1_8h-4  +  24-  2111 

-|.1_7_    3-1-21 
1_7_3-|-21        0 
Thus  1  is  a  root  of  (1),  and  the  quotient  of  the  equation  by 
"^-^^^  a;»-7x='-3a;-f  21  =  0.  (2) 


THEORY  OF  EQUATIONS  179 

If  this  equation  has  any  integral  root  it  must  be  a  factor  of  21. 
We  try  +  3  by  synthetic  division, 

1_7_    3  +  21|3 

+  3  _  12  -  45 
1-4-15-24 

Thus  3  is  not  a  root.    We  try  +  7, 

1_7_3  +  21|7 
+  7  4-  0  -  21 


1+0-3        0 
Thus  7  is  a  root,  and  the  remaining  roots  of  (1)  are  the  roots  of 
a;2  -  3  =  0, 
that  is,  X  =±  V3. 

Hence  the  roots  of  (1)  are  +  1,  +  7,  ±  V3. 

EXERCISES 

Solve  by  trial : 

1.  x3  -  7x2  +  50  =  0.  2,  x3  -  9x  4-  28  =  0. 

3.  x3-36x- 91  =  0.  4.  x3  +  9x  +  26=:0. 

5.  x8  -  19x  +  30  =  0.  6.  x3  -  27x  -  54  =  0. 

7.  x8  +  2x2  -  23x  +  6  =  0.  8.  x3  -  6x2  +  11 X  -  6  =  0. 

9.  x3  -  2x2  -  llx  +  12  =  0.  10.  x3  -  8x2  ^  19a;  _  20  =  0. 

11.  x8  +  9x2  +  27x  +  26  =  0.  12.  x*  -  8x3  +  8x2  ^  40x  -  32  =  0. 

13.  X*  -  13x2  +  48x  -  60  =  0.  14.  x*  -  3x3  -  34x2  +  18x  +168  =  0. 

15.  X* +8x3- 7x2 -50x  + 48  =  0. 

16.  x*  -  3x3  -  5x2  +  29x  -  30  =  0. 

17.  x*  -  6x3  +  13x2  -  30x  +  40  =  0. 

18.  x4-8x3+21x2-34x  +  20  =  0. 

19.  x*  -  12x3  +  43x2  -  42x  +  10  =  0. 

173.  Properties   of  binomial   surds.      A   binomial  surd  is   a 

number  of  the  form  a  ±  V^,  where  a  and  b  are  rational  numbers, 
and  where  b  is  positive  but  not  a  perfect  square. 

Though  we  have  not  explicitly  defined  what  we  mean  by  the  sum  of  an 
irrational  number  and  a  rational  number,  we  shall  assume  that  we  can 
operate  with  the  binomial  surd  just  as  we  would  be  able  to  operate  if  b  were 
a  perfect  square. 


180  ADVANCED  ALGEBRA 

Theorem  \.  If  a  binomial  surd  a  +  V^  =  0^  then  a  =0  anc 
b  =  0. 

J,i  a-\-  V^  =  0  and  either  a  =  0  ov  b  =  0,  clearly  both  must  equal 
zero.  Suppose,  however,  that  neither  a  nor  b  equals  zero.  Then 
transposing  we  have  a  =^  'Vb,  and  a  rational  number  would  be 
equal  to  an  irrational  number,  which  cannot  be.  Hence  the  only 
alternative  is  that  both  a  and  b  equal  zero. 

Theorem  II.  If  two  binomial  surds,  asa  -\-  -sib  and  c  +  V5,  are 
equal,  then  a  =  c  and  b  =  d. 

Let  a  +  V^  =  c  -f  V^. 

Transposing  c,         a  —  c  -\-  V^  =  V5.  (1) 

Square  and  we  obtain 

(^ci  -  cy-\-b  -\-  2(a  -  c)Vb  =  dy 
or 

(^a,  -  cy  +  b  -  d  -\-  2(a  -  c)  Vb  =  0. 

Thus,  by  Theorem  I,  either  b  =  0,  which  is  contrary  to  the  defi- 
nition of  a  binomial  surd,  or  a  ~  e  =  0,  that  is,  a  =  c.  In  the 
latter  case  (1)  reduces  to  Vb  ==  Vd,  OTb  =  d,  and  we  have  a  =  c  and 
b  =  dy  which  was  to  be  proved. 

a  4-  Vb  and  a  —  V^  are  called  conjugate  binomial  surds. 

Theorem  III.  If  a  given  binomial  surd  a  +  Vb  is  the  root  of 
an  equation  with  rational  coefficients,  then  its  conjugate  is  also  a 
root  of  the  same  equation. 

The  proof  of  this  theorem,  which  should  be  performed  in  writ- 
ing by  each  student,  may  be  made  analogously  to  the  proof  of 
the  theorem  on  p.  174. 

174.  Formation  of  equations.  If  we  know  all  the  roots  of  an 
equation,  we  may  form  the  equation  in  either  one  of  two  ways 
(see  p.  167  and  p.  177). 

First  method.  If  a^,  oc^,-",  oc^  are  the  given  roots,  multiply 
together  the  factors  x  —  a^,- ■•,x  —  a^. 

Second  method.  From  the  given  roots  form  the  coefficients 
by  the  rule  on  p.  177. 


THEORY  OF  EQUATIONS  181 

If  the  equation  and  all  but  one  of  its  roots  are  known,  that 
root  can  be  found  by  the  solution  of  a  linear  equation  obtained 
from  the  coef&cient  of  the  second  or  the  last  term.  If  all  but  two 
of  its  roots  are  known,  the  unknown  roots  may  be  found  by  the 
solution  of  a  pair  of  simultaneous  equations  formed  from  the 
same  coefficients. 

In  the  solution  of  the  following  exercises  use  is  made  of  the 
theorem  on  p.  174,  Theorem  III,  p.  180,  and  the  various  relations 
between  the  roots  and  the  coefficients. 

EXERCISES 

1.  Form  the  equations  which  have  the  following  roots.  Check  the  process 
by  using  both  methods  of  §  174. 

(a)  2,  -  3,  1. 
Solution : 

First  metfiod.  {x  -  2){x  +  3)  (x  -  1)  =  JC^  -  7  x  +  6. 

Second  method.   Let  the  equation  be 

x3  +  6ix2  +  biX  +  63  =  0. 
Then,  by  §  170,  61  =  -  (2  -  3  +  1)  =  0,  -  •  \ 

62  =  -6  +  2-3  =  -7, 

63  =  -2-3. -1  =  6. 
The  equation  then  is  x^  —  7x  +  6  =  0. 

(b)  1,  2,  3.  (c)  2,  2,  2,  2. 
(d)  3,  1,  1,  0.  (e)  1,  0,  0,  0. 
(f)  ±V2,±i.  (g)  2,4,  -6. 
(h)  2,  -  3,  1,  0.  (i)  2,  3,  -  6. 

.     (j)  7,  V5,  -V5.  (k)  1,2,  -1,  -|. 

(1)  3,  1  +  i,  1  -  i  (m)  _  4,  -  3,  3  ±  V5. 

(n)  1  ±  i,  -  1  ±  i  (o)  2,  V=^,  -V^^. 

(p)   -1,2,3,-4.  (q)2^,3|,  -H,  -2^. 
(r)  ±V6,  ±iV7.  (s)   -5,  2  +  V6,  2-V5. 

(V)  3,  ^i±^,  i:!^.         (w)  - 1,  l±i^^  1^. 


182 


ADVANCED  ALGEBRA 


2.  The  equation  x*  +  2x3  -  7^2  _  8x  +  12  =  0  has  two  roots  —  3  and 
+  1.    Find  the  remaining  roots. 

Solution :  Let  the  unknown  roots  be  a  and  h. 
Then,  by  §  170,  _a-6  +  3-l  =  2, 

-  3  a6  =  12. 
Solving  for  a  and  5,  we  obtain  a  =  —  2  or  +2, 

6  =  +  2  or  -  2. 

3.  x8  —  7x  +  6  =  0  has  the  roots  2  and  1.    Find  the  remaining  root. 

4.  X*  —  3x  +  2  =  0  has  the  root  1.    Find  the  remaining  roots. 

5.  x^  —  18  X  —  35  =  0  has  the  root  5.    Find  the  remaining  roots. 

6.  Two  roots  of  x*  —  35  x^  -f-  90  x  —  56  =  0  are  1  and  2.  Find  the  remain- 
ing roots. 

7.  The  roots  of  x^  -  6  x2  -  4  x  +  24  =  0  are  in  A.  P.    Find  them, 

8.  The  two  equations  x3-6x2  +  llx-6  =  0  and  x^  -14x2  +  63  x  -  90  =  0 
have  a  root  common.  Plot  both  equations  on  the  same  axes,  and  find  all  the 
roots  of  both  equations. 

9.  Determine  the  middle  term  of  the  equation  whose  roots  are  —  2, 
+  1,  3,  —  4  without  determining  any  other  term. 

10.  What  is  the  last  term  of  the  equation  whose  roots  are  —  4,  4,  ±  V—  3  ? 

11.  One  root  of  x*  -  4x8  +  5x2  +  2x  +  52  =  0  is  3  -  2i.    Find  the  remain- 
ing roots. 

12.  One  root  of  x*  -  4  x^  +  5  x2  +  8  x  - 14= 0  is  2  +  i  V3.    Find  the  others. 

13.  Plot  the  following  equations,  determine  all  the  integral  roots,  and 
find  the  remaining  roots  by  solving. 

(a)  X*  -  6x8  +  24x  -  16  =  0. 


X 

y 

-1 

-   33 

-2 

0 

-3 

+  155 

In  this  plot  two  squares  on  the  X  axis  represent  a  unit  of  x,  while  one 
square  on  the  Y  axis  represents  ten  units  of  y.  The  integral  factors  are 
X  —  2  and  x  +  2,  since  ±  2  are  roots,  that  is,  are  values  of  x  for  which  the 


THEORY  OF  EQUATIONS  183 

curve  is  on  the  X  axis.  To  find  the  quotient  of  our  equation  we  first  divide 
synthetically  by  2,  and  then  the  quotient  by  —  2,  using  the  principle  given 
in  §  161. 

1-6+    0  +  24- 16 12 
+  2  -    8-16  +  16 
^  1-4-    8+    8        0|-2 

-  2  +  12  -    8 
1-6+    4        0 

Thus  the  quotient  of  the  polynomial  and  (x  —  2)  («  +  2)  is  «2  __  ga;  4.  4 
Solving  the  equation 

x2-6x  +  4  =  0, 

we  obtain  the  two  remaining  roots,  x  =  3  ±  V5.  These  remaining  roots 
might  also  be  found  by  the  method  of  exercise  2. 

(b)  x3  -  6x  -  12  =  0.  (c)  x3  -  8x2  +  7  =  0. 

(d)  x8  -  7x2  +  50  =  0.  (e)  x^  -  8x2  +  13x-  6  =  0. 

(f)  x8-6x2+7x-2  =  0.  (g)  x3  + 3x2  + 4x- 24  =  0. 

(h)  X*  -  3x3  +  7 x2  -  21 X  =  0.  (i)  X*  -  3x3  -  7 x2  +  27 x  -  18  =  0. 

'(j)  x*-9x3  +  21x2-19x  +  6  =  0. 

(k)  How  many  imaginary  roots  can  an  equation  of  the  5th  degree  have  ? 
(1)  x3  —  ax2  +  6x  +  c  =  0  has  two  roots  whose  sum  is  zero.    What  is  the 
third  root  ?    What  are  the  two  roots  whose  sum  is  zero  ? 

(m)  x3  +  x2  +  6x  +  c  =  0  has  one  root  the  reciprocal  of  the  other.    What 
are  the  values  of  the  roots  ? 

(n)  x3  —  4x2  +  ox  +  62  =  0  has  the  sum  of  two  roots  equal  to  zero.    What 
must  be  the  values  of  a  and  h  ? 

(o)  X*  -  3  x3  +  6x  +  9  =  0  has  the  sum  of  three  of  its  roots  equal  to  zero. 
What  must  be  the  value  of  6  ? 

175.  To  multiply  the  roots  by  a  constant.    Suppose  we  have 
the  equation 

fix)  =  a,x-  +-  a^x--'  +  . . .  +  «^  =  0,  (1) 

whose  roots  are  «ri,  a^,  •  •  • ,  «:„.  An  equation  of  this  type  for 
values  of  n  greater  than  2  is  usually  not  solvable  by  elementary 
methods.  It  often  happens,  however,  that  by  changing  its  form 
slightly  we  may  obtain  an  equation  one  or  more  of  whose  roots 
we  can  find.  We  shall  see  that  if  an  equation  has  rational  roots 
we  may  always  find  them  if  we  change  the  form  of  the  equation 
as  indicated  on  the  following  page. 
/ 


184  ADVANCED  ALGEBRA 

We  seek  to  form  from  (1)  an  equation  whose  roots  are  equal 
to  the  roots  of  (1)  multiplied  by  a  constant  factor,  as  k.  Thus 
the  equation  we  seek  must  have  the  roots  kai,  ka^,  ka^.  We 
carry  out  the  proof,  which  is  perfectly  general,  on  the  equation 
of  the  third  order 

f(x)  =  a^x^  -f-  a-^x^  -f  «2^  +  ^3  =  0, 

whose  roots  are  a^,  a^,  a^.  The  equation  that  we  seek  must  have 
roots  kai,  ka^,  ka^.  Since  now  (§  53)  f{x)  =  0  is  satisfied  by 
a,  where  a  stands  for  any  one  of  the.  roots,  that  is,  since  f(a)  =  0, 

evidently  /(t)  =  0  is  satisfied  by  ka,  that  is. 

Hence  we  obtain  an  equation  that  is  satisfied  by  ka^,  ka^^  ka^, 
if  in/(£c)  we  let  a;  =  t* 

The  required  equation  is  then 

Akj-'W^l^^-k^''^-^^ 

or,  multiplying  by  k^, 

a^z^  +  ka-^z'^  -{-  k^azZ -{-  k^a^  —  0. 
This  affords  the  general 

KuLE.  To  multiply  the  roots  of  an  equation  hy  a  constant  k, 
multiply  the  successive  coefficients  beginning  with  the  coefficient  of 
x""'^  hy  k,l^,  •  •  -,1^  respectively. 

In  performing  this  operation  the  lacking  powers  of  x  should  be 
supplied  with  zero  coefficients. 

Example.   Multiply  the  roots  of  2  x^  —  3  x  +  4  =  0  by  2. 
Multiply  the  coefficients  by  the  rule  above, 

2«8  +  2-0x2-4-3x  +  8-4  =  0. 
Simplifying,  »»  -  6  x  +  16  =  0. 


THEORY  OF  EQUATIONS  185 

When  an  equation  in  form  (2),  p.  166,  has  fractional  coefficients, 
an  equation  may  be  formed  whose  roots  are  a  properly  chosen 
multiple  of  the  roots  of  the  original  equation  and  whose  coeffi- 
cients are  integers. 

Corollary  I.  When  k  is  a  fraction  this  method  serves  to 
divide  the  roots  of  an  equation  hy  a  given  number. 

Corollary  IL  When  k  =  —  1  this  method  serves  to  form  an 
equation  whose  roots  are  equal  to  the  roots  of  the  original  equa- 
tion hut  opposite  in  sign.  This  is  equivalent  to  the  statement 
that  /(—  x)  =  0  has  roots  equal  hut  opposite  in  sign  to  those 

•  EXERCISES 

1.  Form  the  equation  whose  roots  are  three  times  the  roots  of 

Solution :  Supplying  the  missing  term  in  the  equation,  we  have 

«* -6x3  +  0x2- X +  1  =  0. 
Since  A;  =  3,  we  liave  by  the  rule 

x* -3 -6x3 +  9 -0x2- 27 -x  +  81  =  0,  'r 

or  x*-18x3 -27x  +  81  =  0.  ' 

2.  Find  the  equation  whose  roots  are  twice  the  roots  of 

x*  +  3x3-2x  +  4  =  0, 

3.  Find  the  equation  whose  roots  are  one  half  the  roots  of 

x3-2x2  +  3x-4  =  0. 

4.  Find  the  equation  whose  roots  are  two  thirds  the  roots  of 

x3-4x-6  =0. 

5.  By  what  may  the  roots  of  the  following  equations  be  multiplied  so  thai 
in  the  resulting  equation  the  coefficient  of  the  highest  power  of  x  is  unity 
and  the  remaining  coefficients  are  integers  ?    Form  the  equations. 

(a)  3x3-  6x  +  2  =  0. 

Solution :  We  wish  to  bring  into  every  term  such  a  factor  that  all  the 
resulting  coefficients  are  divisible  by  3. 

Let  A:  =  3. 

Supply  the  lacking  term, 

3x3  +  0x2  -6x  +  2  =  0. 

By  rule,  3x3  +  3  •  0x2  -  9  •  6x  +  27  •  2  =  0. 

•  Dividing  by  3,  x3  -  18  x  +  18  =  0. 


186 


ADVANCED  ALGEBRA 


x2 

(b)    X3  +  -  - 


1  =  0. 


(C)   X8  -  i  =  0. 


(d)  «»  +  7x2  +  -x  + -  =  0, 


62 


6» 


(e)  x4  +  ^ 


x2 


+  1  =  0. 


(f)  2x8  -  3x2 -x  + 4  =  0. 
(h)  x*-6x3-2x2+  1  =0. 


(g)  3x4-3x2-4x  +  l  =  0. 
(i)  16x4  -  24x3  +  8x2 -2x  + 1  =  0. 


6.  Form  equations  whose  roots  are  the  negatives  of  the  roots  of  the  fol- 
lowing equations. 

(a)  x8-4x  +  6  =  0. 

Solution :  Supply  the  lacking  term, 

x3+  0x2-4x  +  6  =  0. 
Changing  signs  we  obtain  by  Corollary  II 

x3_0x2-4x-  6  =  0, 
or  x3  -  4  X  -  6  =  0. 

(b)  X'  -  2x2  -  4x  =  0.  (c)  x*  -  3 x2  +  1  =  0. 

(d)  X*  -  2x3  +  x2  4-  2x  -  1  =  0.  (e)  x3  +  3x2  +  7x  -  13  =  0. 

7.  What  effect  does  changing  the  sign  of  every  term  of  the  member 
involving  x  have  on  the  graph  of  an  equation? 

8.  What  is  the  graphical  interpretation  of  the  transformation  which 
changes  the  signs  of  the  roots  of  an  equation,  that  is,  what  relation  does 
the  graph  of  the  equation  before  transformation  bear  to  the  graph  of  the 
equation  after  transformation  (a)  when  the  degree  is  an  even  number, 
(b)  when  the  degree  is  an  odd  number? 

9.  If  4x4  -  16x3  -  86x2  +  4x  +  21  =  0  has  as  two  roots  -  ^  and  - 3, 
what  are  the  roots  of  4x4  +  16x3  -  85x2  -  4x  +  21  =  0? 

10.  If  a  root  of  xs  -  11  x2  +  36  x  -  36  =  0  is  2,  what  are  the   roots  of 
x»+ 11x2+ 36x  + 36  =  0? 


176.  Descartes'  rule  of  signs.  A  pair  of  successive  like  signs  in 
an  equation  is  called  a  continuation  of  sign.  A  pair  of  successive 
unlike  signs  is  called  a  change  of  sign. 


In  the  equation 


2x4_3a;8  +  2x2  +  2a;-3  =  0 


(1) 


are  one  continuation  of  sign  and  three  changes  of  sign.    This  may  be  seen  more 
clearly  by  writing  merely  the  signs,  +   —   +   +   —. 

Let  US  now  inquire  what  effect  if  any  is  noted  on  the  number 
of  changes  of  sign  in  an  equation  if  the  equation  is  multiplied  by 


THEORY  OF  EQUATIONS  187 

a  factor  of  the  form  x  —  a  when  a  is  positive,  that  is,  when  the 
number  of  positive  roots  of  the  equation  is  increased  by  one. 
Let  us  multiply  equation  (1)  by  a;  —  2.    We  have  then 

x-2 

2x^  -3x*-{-2x^-{-2x''-Sx 

-4:X^+6x^-4:X^-4:X-\-6 

2x'  -  7  x^  -{-  Sx^  -  2x^  -  7  x  -{-  6 

In  this  expression  the  succession  of  signs  is  +  —  +  —  —  +, 
in  which  there  are  four  changes  of  sign,  that  is,  one  more  change 
of  sign  than  in  (1).  If  an  increase  in  the  number  of  positive  roots 
always  brings  about  at  least  an  equal  increase  in  the  number  of 
changes  of  sign,  there  must  be  at  least  as  many  changes  of  sign  in 
an  equation  as  there  are  positive  roots.  This  is  the  fact,  as  we 
now  prove. 

Descartes'  rule  of  signs.  An  equation  f(x)  =  0  has  no  more 
real  positive  roots  than  f(x)  has  changes  of  sign. 

Illustration.  In  the  equation  of  degree  one  £c  —  2  =  0  there 
is  one  change  of  sign  and  one  jjositive  root.  In  the  case  of  a  linear 
equation  there  is  no  possibility  of  more  than  one  change  of  sign. 
In  the  quadratic  equation  x'^-\-2x-\-l  =  Q  there  is  no  change  of 
sign,  and  also  no  positive  root  since  for  positive  values  of  x  the 
expression  x^  -f  2  cc  +  1  is  always  positive  and  hence  never  zero. 
In  the  equation  ic^  +  2ic  —  3  =  0  we  have  one  change  of  sign, 
and  one  positive  root,  + 1. 

We  shall  prove  this  general  rule  by  complete  induction. 

First.  W^e  have  just  seen  that  the  rule  holds  for  an  equation 
of  degree  one. 

Second.  We  assume  that  the  rule  holds  for  an  equation  of 
degree  m,  and  prove  that  its  validity  for  an  equation  of  degree 
m  -\-l  follows.  We  shall  show  that  if  we  multiply  an  equation 
of  degree  m  hj  x  —  a,  where  a  is  positive,  thus  forming  an  equa- 
tion of  degree  m  + 1,  the  number  of  changes  of  sign  in  the  new 
equation  always  exceeds  the  number  of  changes  of  sign  in  the 


188 


ADVANCED  ALGEBRA 


original  equation  by  at  least  one.    That  is,  the  number  of  chanj 
of  sign  increases  at  least  as  rapidly  as  the  increase  in  the  number 
of  positive  roots  when  such  a  multiplication  is  made. 

Let/(£c)  =  0  represent  any  particular  equation  of  the  nth.  degree. 
The  first  sign  of  f(x)  is  always  +.  The  remaining  signs  occur  in 
successive  groups  of  +  or  —  signs  which  may  contain  only  one 
sign  each.  If  any  term  is  lacking,  its  sign  is  taken  to  be  the  same 
as  an  adjacent  sign.  Thus  the  most  general  way  in  which  the 
signs  of  /(x)  may  occur  is  represented  in  the  following  table, 
in  which  the  dots  represent  an  indefinite  number  of  signs.  The 
multiplication  of  f(x)  by  a;  —  a  is  represented  schematically,  onl^P"-^ 
the  signs  being  given.  ^ 


fix) 

X  —  a 

All  +  signs 
+   ••••   + 

All  -  signs 

All  +  signs 

All  -  signs 

Further 
groups 

+   ••••  + 

All  -  signs 

+         - 

xf{x) 
-  ocfix) 

+  +  •••  + 

-  +  •••  + 

+  +  •••  + 
+ 

-  +  •••  + 

+  +•••  + 
+ 

-+         +  + 

{x-a)f{x) 

+  d=---± 

-±.-.± 

+  ±---± 

-±...± 

+  ±         ± 

-±•••±4- 

The  ±  sign  indicates  that  either  the  +  or  the  —  sign  may  occur 
according  to  the  value  of  the  coefficients  and  of  a.  The  verti- 
cal lines  denote  where  changes  of  sign  occur  in  f(x).  Assuming 
that  all  the  ambiguous  signs  are  taken  so  as  to  afford  the  least 
possible  number  of  changes  of  sign,  even  then  in  (x  —  cc)f(x) 
there  is  a  change  of  sign  at  each  or  between  each  pair  of  the 
vertical  lines,  and  in  addition,  one  to  the  right  of  all  the  vertical 
lines.  Thus  as  we  increase  the  number  of  positive  roots  by  one 
the  number  of  changes  of  sign  increases  at  least  by  one,  perhaps 
by  more. 

The  only  possible  variation  that  could  occur  in  the  succession 
of  groups  of  signs  in  f(x),  namely,  when  the  last  group  is  a 
group  of  4-  signs,  does  not  alter  the  validity  of  the  theorem. 


THEORY  OF  EQUATIONS  189 

We  illustrate  the  foregoing  proof  by  the  following  particular 
example. 

Let  f{x)  =  x^  -4:X^  -  x-{-2,  and  let  a  =  2. 

Multiply  f(x),     1  +  0-4  +  0-1  +  2  4  changes 

by  x-2,    1-2 

xf(x),  1  +  0-4  +  0-1  +  2 

-2f(x),  -2-0  +  8-0  +  2-4 

(x~2)f(x),  1-2-4  +  8-1  +  4-4                 5  changes 

177.  Negative  roots.  Since /(—a:)  has  roots  opposite  in  sign 
to  those  of  f(x)  (p.  185),  we  can  state 

Descartes'  rule  of  signs  for  negative  roots.  f(x)  has  nc 
more  negative  roots  than  there  are  changes  in  sign  in  /(—  x). 

If  by  Descartes'  rule  it  appears  that  there  cannot  be  more  than 
a  positive  roots  and  b  negative  roots,  and  if  a  -{-  b  <n,  the  degree 
of  the  equation,  then  there  must  be  imaginary  roots,  at  least 
n  —  (a  -{-  b)  in  number. 

EXERCISES 

1.  Prove  Descartes'  rule  of  signs  for  x^  +  bx  +  c  =  0  directly  from  the 
expression  for  b  and  c  in  terms  of  the  roots  (see  §  115). 

2.  Find  the  maximum  number  of  positive  and  negative  roots  and  any 
possible  information  about  imaginary  roots  in  the  following  equations. 

(a)  x3  +  2  x2  +  1  =  0. 

Solution :  Writing  signs  of  f{x),  +  +  +,  there  is  no  change,  hence  no 
positive  root. 

Writing  signs  of  /(— £c),  — +  +  ,  there  is  one  change,  hence  no  more 
than  one  negative  root.  Since  there  can  be  only  one  real  root  there  must 
be  two  imaginary  roots. 

(b)  x3  +  1  =  0.  (c)  X*  -  2  =  0. 

(d)  x3  -  X  +  1  =  0.  (e)  x6  -  X  +  1  =  0. 

(f )  x4  +  X  +  1  =  0.  (g)  x6  +  x2  +  1  =  0. 

(h)  x3-6x2+4x-l  =  0.  (i)  x5-2x4-3x3+4x2+x  +  l  =  0. 

(i)  x6  +  2x*  -  6x3  -  4x2  +  X  -  1  =  0. 


190  ADVANCED  ALGEBRA 

178.  Integral  roots.  In  finding  the  rational  roots  of  an  equa- 
tion we  make  use  of  the  following 

Theorem.    If  the  equation 

^  +  «i^"-'+---+a„  =  ^  (1) 

{where  the  oUs  are  integers)  has  any  rational  root,  such  root  must 
he  an  integer. 

Suppose  -  be  a  fraction  reduced  to  its  lowest  terms  which 
satisfies  the  equation. 

Then  ^  +  2l£r!  +  ...  +  „„  =  0 

is  an  identity. 

Then  clearing  of  fractions  and  transposing, 

f  =  --q  {ciip"~^  H h  a„!?"~0- 

Thus  some  factor  of  g'  is  a  factor  of  ^^  that  is,  oip  (p.  52),  which 
contradicts  the  hypothesis  that  -  is  reduced  to  its  lowest  terms. 

Thus  all  the  rational  roots  of  the  equation  are  integers,  which 
as  we  know  (§  170)  are  factors  of  a„. 

179.  Rational  roots.   If  we  seek  the  rational  roots  of 

aox""  H [-  a„  =  0, 

where  a^  =^  1,  we  can  multiply  the  roots  by  a  properly  chosen 
constant  (§  175)  and  obtain  an  equation  of  form  (1)  above  whose 
integral  roots  may  easily  be  found  by  synthetic  division. 

Example.   What  rational  roots,  if  any,  has 

8x»+lla;-14  =  0?  (1) 

Multiply  the  roots  by  8,    3  a^  +  99  x  -  378  =  0. 

Divide  by  3,  x^  +  33  x  -  126  =  0.  (2) 

Since  by  Descartes'  rule  of  signs  equation  (1)  has  only  one  positive  root 
and  no  negative  root, we  do  not  need  to  carry  the  table  further  than  to  test 
for  a  positive  root. 


THEOKY  OF  EQUATIONS  191 

y  Form  a  table  of  values  for  equation  (2)  by  synthetic  division. 

~^  We  need  only  to  try  the  factors  of  126  (§  170). 
_  g2  Thus  (2)  has  the  root  3.    Hence  the  original  equation  (1)  has 

Q  the  root  3^3  =  1. 


KuLE.  To  find  all  the  rational  roots  of  an  equation,  trans- 
form the  equation  so  that  the  first  coefficient  is  + 1. 

Find  the  maxiymum  number  of  positive  and  negative  roots  by 
Descartes'  rule  of  signs. 

Find  the  integral  roots  of  this  equation  by  trial,  and  the 
roots  of  the.  original  equation  by  dividing  the  integral  roots 
found  by  the  constant  by  which  the  roots  were  multiplied. 

By  the  Theorem  §  178  we  are  assured  that  all  the  rational 
roots  can  be  found  in  this  way. 

EXERCISES 

Find  all  the  rational  roots  of  the  following  equations. 

1.  4x8  =  27(x  +  l).  2.  15jc8  +  13a;2-2  =  0. 

3.  4a;3-6x-6  =  0.  4.  x^- 2f  x2  + 2|x  -  1  =  0. 

5.  4x3  -  8x2  -  X  +  2  =  0.  6.  3x*  -  8x8  _  ^q^2  +  25  =  0. 

7.  4x8-4x2+ x-6  =  0.  8.  3x8+ 13x2+ llx  -  14  =  0. 

9.  4x8  +  16x2 -9x- 36  =  0.  10.  2x3  -  21x2  +  74x  -  85  =  0. 

11.  6x8  -  47x2 +  71X  + 70  =  0.  12.  12x8  -  52x2+ 23x  +  42  =  0. 

13.  6x8-29x2+ 53x- 45  =  0.  14.  6x*- x8- 8x2- 14x  +  12  =  0. 
15.  27x3+ 63x2+ 30x- 8  =  0.  16.  2x*-13x8  +  16x2- 9x +20  =  0. 
17.  3x8  -  26x2 +52x- 24  =  0.  18.  Ox*  -  x3  -  49x2+ 55x  -  50  =  0. 
19.   18x8+ 81x2+ 121x +  60  =  0.       20.   12x4  +  6x8  -  24x2- 9x  +  9  =  0. 

21.  10x4  +  18x8  -  16x2 +  8x- 20  =  0. 

22.  9x4  +  15x8  -  143x2 +41X  + 30  =  0. 

23.  36x4  -  72x8  -  31  x2  +  67  x  + 30  =  0. 

24.  24x4- 108x8+ 324x2 -240X  + 60  =  0. 

180.  Diminishing  the  roots  of  an  equation.  In  the  preceding 
sections  we  have  solved  completely  the  problem  of  finding  the 
rational  roots  of  an  equation.    We  now  pass  to  the  problem  of 


V 


192  ADVANCED  ALGEBRA 


finding  the  approximate  values  of  the  irrational  roots  of  an  eqnar 
tion.  In  carrying  out  the  process  that  we  shall  develop  it  is 
desirable  to  form  an  equation  whose  roots  are  equal  respectively 
to  the  roots  of  the  original  equation  each  diminished  by  a  constant. 

Let  fix)  =  aox""  +  a^x''-'^  -\ h  ««  =  0,  (1) 

whose  roots  are  ^i,  a^,  •  ••,  a^.  Let  a  be  any  constant.  We  seek 
an  equation  whose  roots  are  a^  —  a,  a^  —  a,  -  •  • ,  a^  —  a.  ^~^^^v_ 

If  we  let  a  stand  for  any  one  of  the  roots  of  (1),  since  f(a)  =  0 
(p.  33),  we  see  that 

/(«  +  a)  =  0  is  satisfied  by  a  —  a, 
that  is,  f{a-a-\-a)=  f(a)  =  0. 

Thus  to  form  the  desired  equation  replace  x  by  z  -\-  a.    We 
obtain 

f(x)  =  f(z  +  a)  =  ao(z  +  a)»  ■}- a,{z -{-  af-^  +  . . .  +  a„  =  0. 

Developing  each  term  by  the  binomial  theorem  and  collecting 
like  powers  of  z,  we  get  an  equation  of  the  form 

fix)  =  Fiz)  =  A,z-  +  A^z-^  +  . . .  +  ^^  =  0,  (2) 

where  the  ^'s  involve  the  a's  and  d.  This  is  the  equation  desired. 
We  now  seek  a  convenient  method  of  finding  the  values  of  the 
coefficients  ^o?  ^i?  ^2>  •••>  ^n  when  a^^  %,  a^^  •••,  a„  are  given 
numerically.  Now  A^  is  the  remainder  from  the  division  of  Fiz) 
by  z.  But  since  Fiz)  =fix)  and  z  =  x  —  a,  the  remainder  from 
dividing  Fiz)  by  z  is  identical  with'  the  remainder  from  dividing 
fix)  hj  x  —  a.  Thus  A^  is  the  remainder  from  dividing  fix)  by 
X  —  a.    Furthermore,  since  A^_y^  is  the  remainder  from  dividing 

—^^ by  z,  it  is  also  the  remainder  from  dividing  ^^-^^ 

by  a;  —  a.  The  process  may  be  continued  for  finding  the  other  ^'s. 
We  may  then  diminish  the  roots  of  an  equation  by  a  as  follows : 

EuLE.    The  constant  term  of  the  new  equation  is  the  remaivr- 
der  from  dividing  fix)  hy  x  —  a. 


i 


THEORY  OF  EQUATIONS 


193 


The  coefficient  of  z  in  the  new  equation  is  the  remainder  from 
dividing  the  quotient  just  obtained  hy  x  —  a. 

The  coefficients  of  the  higher  powers  of  z  are  the  remainders 
from  dividing  the  successive  quotients  obtained  hy  x  —  a. 

Example.  Form  the  equation  whose  roots  are  2  less  than  the  roots  of 
x4  _  2  x3  -  4  x2  4-  x  -  1  =  0. 

The  divisions  required  by  the  rule  we  carry  out  synthetically  (p.  169). 
1-2-4+1-    1[2 
+  2  +0  -8  -14 
-15 


1  +  0 

+  2 

-4  -7 

+  4  +0 

1  +  2  +0 
+  2  +8 

-7 

1  +  4 

+  2 

+  8 

1+6 

The  desired  equation  is 

^4  +  6x3  +  8x2 


7x-15  =  0. 


181.  Graphical  interpretation  of  decreasing  roots.  If  an  equa- 
tion has  roots  a  units  less  than  those  of  another  equation,  if  a  is 
positive  its  intersections  with  the  X  axis  or  with  any  line  parallel 
to  the  X  axis  are  a  units  to  the  left  of  the  corresponding  inter- 
sections of  the  first  equation.  It  is,  in  fact,  the  same  curve,  except- 
ing that  the  Y  axis  is  moved  a  units  to  the  right.  If  a  is  negative, 
the  Y  axis  is  moved  to  the  left. 


EXERCISES 

Plot,  decrease  the  roots  by  a  units,  and  plot  the  new  axes. 

1.  x4-3x3 -2x-3  =  0.    a  =  3. 

Solution:  1-3-    0-    2-3[3 


(1) 


+  3-0 

-    0 

-6 

1_0  -    0  -    2 
+  3  +    9+27 

-9 

1  +  3  +    9 

+  3  +  18 

+  25 

1  +  6 

+  3 

1  +  9 


+  27 


194 


ADVANCED  ALGEBRA 


Thus  the  required  equation  is 


0.        (2) 


X 

y 

0 

-    3 

:  1 

-    7 

2 

-16 

3 

-    9 

-1 

+    3 

3.  x3  _  8  =  0.    a  =  1.4. 
5.  x3  +  4x-8  =  0.    a  =  3. 
7.  x3  +  2 X  +  6  =  0.    a  =  -\. 
9.  «8-2x2  +  8x-7  =  0.    a  =  2. 

11.  x4-3x2  +  2x 

12.  X 

13.  2x3-6x2 +  4x -3  =  0 

14.  X*  +  6x3  +  10x2  +  x  -  1  =  0 


In  the  figure  one  square  on  the  Faxis  repre- 
sents two  units  of  y,  and  two  squares  on  the 
X  axis  represent  one  unit  of  x. 

2.  x*-16  =  0.    a  =  2. 
4.  x4- 2x2  +  1  =  0.    a  =  .2. 
6.  x3-4x2-2  =  0.    a  =  .5. 
8.  x3  +  4  x2  +  X  -  6  =  0.    a  =  -  .4. 
10.  x3-3x2  +  x-l  =  0.    a  =  -.3. 
2  =  0.    a  =  -2. 
15x2 +  7x  + 125  =  0.    a  =  5.  - 

a  =  -3. 

a  =  -l. 


182.  Location  principle.  If  when  plotting  an  equation  y  =f(x) 
the  value  x  =  a  gives  the  corresponding  value  of  y  positive  and 
equal  to  c,  while  the  value  x  =  b  gives  the  corresponding  value 
of  y  negative,  say  equal  to  —  d,  then  the 
point  on  the  curve  x  =  a,  y  =  c  is  above 
the  X  axis,  and  the  point  on  the  curve 
x  =  byy  =  —  -^  is  below  the  X  axis.  If  our 
curve  is  unbroken,  '^  must  then  cross  the 
X  axis  at  least  once  between  the  values 
X  —  a  and  x  =  h^  and  hence  the  equation 
must  have  a  root  between  those  values 
of  x.  The  shorter  we  can  determine  this 
interval  a  to  b  the  more  accurately  we  can  find  the  root  of  the 
equation.  This  property  of  unbrokenness  or  continuity  of  the 
graph  of  ?/  =  ^0^"  4-  aix"~^  +  •  •  •  +  a„  we  assume.  We  assume 
then  the  following 


THEORY  OF  EQUATIONS 


195 


Location  principle.    When  for  two  real  unequal  values  of  x, 
x  =  a  and  x  =  h,  the  value  of  y  —f{x)  has  opposite  signs, 
the  equation  f{x)  =  0  has  a  real  root  between  a  and  h. 

Illustration.    The   equation  f(x)  =ic^-f3cc  —  5  =  0   has  a 
root  between  1  and  2.    Since  /(I)  =  -  1,  /(2)  =  9. 

183.  Approximate  calculation  of  roots  by  Horner's  method. 

We  are  now  in  a  position  to  compute  to  any  required  degree  of 

accuracy  the  real  roots  of  an  equation.    Consider  for  example 

the  equation  ^s  +  3,_20  =  0.  (1) 

Form  the  table  of  values  for  plotting  the  equation 

ic8  +  3  a:  -  20  =  ?/. 

By  the  location  theorem  we  find 
that  a  root  is  between  +  2  and  +  3. 
To  find  more  precisely  the  position 
of  the  root  we  might  estimate  from 
the  graph  the  position  of  the  root 
and  substitute  say  2.3,  2.4,  and  so  3+16 

on,  until  we  found  two  values  be- 
tween which  the  root  lies.  We  can 
gain  the  same  result  with  much  _  i  _  24 
less  computation  if  we  first  dimin- 
ish the  roots  of  the  equation  so  that 
the  origin  is  at  the  less  of  the  two  integral  values  between  which 
we  know  the  root  lies.    Here  we  decrease  the  roots  of  (1)  by  2, 

1  4-  0  +    3  -  20[2 


y 
-20 

-16 


+  2  +    4+14 

1  +  2  +    7 
+  2  -h    8 

-    6 

1  +  4 
+  2 

+  15 

1+6 

The  equation  whose  roots  are  decreased  by  2  is 
a:»  +  6ic2  +  15a;^6  =  0. 


(2) 


196  ADVANCED  ALGEBRA 

We  know  that  (2)  has  a  root  between  0  and  1,  since  equation  (1) 
has  a  root  between  2  and  3.  From  the  graph  we  can  estimate  the 
position  of  the  root.  Having  made  an  estimate,  say  .3,  it  is  neces- 
sary to  verify  the  estimate  and  determine  by  synthetic  divi^io^ 
precisely  between  which  tenths  the  root  lies.    Thus,  trying  .3,  we" 

^^*^^^  1  +  6.0  +  15.00  -  6.000[_^ 

+  0.3+    1.89  +  5.067 
1  +  6.3  +  16.89  -  0.933  () 

which  shows  that  for  x  =  .3  the  curve  is  below  the  X  axis,  hence 
the  root  is  greater  than  .3.  But  we  are  not  justified  in  assuming 
that  the  root  is  between  .3  and  .4  until  we  have  substituted  .4 
for  X.    This  we  proceed  to  do. 

1 +  6.0 +  15.00- 6.000  [^ 

+  0.4+    2.56  +  7.024 
1  +  6.4  +  17.56  +  1.024 

Since  the  value  of  y  is  positive  for  x  =  .4,  the  location  principle 
shows  that  (2)  has  a  root  between  .3  and  .4,  that  is,  (1)  has  a  root 
between  2.3  and  2.4. 

To  find  the  root  correct  to  two  decimal  places,  move  the  origin  up 
to  the  lesser  of  the  two  numbers  between  which  the  root  is  now 
known  to  lie.    The  new  equation  will  have  a  root  between  0  and  .1. 

This  process  is  performed  as  follows : 

1  +  6.0  +  15.00  -  6.000[^- 
+  0.3  +    1.89  +  5.067 


1  +  6.3  +  16.89 
+  0.3  +    1.98 


1  +  6.6 
+  0.3 


0.933 


+  18.87 


1  +  6.9 

Thus  the  new  equation  is 

x^  +  6.9  x^  +  18.87  X  f  .933  =  0.  (3) 

This  equation  has  a  root  between  0  and  .1.    We  can  find  an 
approximate  value  of  the  hundredths  place  of  the  root  by  solving 


THEORY  OF  EQUATIONS  197 

the  linear  equation  18.87  x  —  .933  =  0,  obtained  from  (3)  by  drop- 
ping all  but  the  term  in  x  and  the  constant  term. 

This  suggestion  must  be  verified  by  synthetic  division  to  deter- 
mine between  what  hundredths  a  root  of  (3)  actually  lies. 

1  +  6.90  +  18.870  -  0.9330|.04 
+  0.04  +    0.277  +  0.7658 

6.94  +  19.147  -  0.1672 

Thus  the  curve  is  below  the  X  axis  at  a;  =  .04  and  hence  the 
root  is  greater  than  .04.  We  must  not  assume  that  the  root  is 
between  .04  and  .05  without  determining  that  the  curve  is  above 
the  X  axis  at  x  =  .05. 

1  +  6.90  -f  18.870  -  0.9330[^ 
+  0.05+    0.347  +  0.9608 

6.95  +  19.217  +  0.0278 

Thus  the  curve  is  above  the  X  axis  ai  x  =  .05.  By  the  location 
principle  (3)  has  a  root  between  .04  and  .05,  that  is,  (1)  has  a 
root  between  2.34  and  2.35.  We  say  that  the  root  2.34  is  correct 
to  two  decimal  places.  If  a  greater  degree  of  precision  is  desired, 
the  process  may  be  continued  and  the  root  found  correct  to  any 
required  number  of  decimal  places. 

The  foregoing  process  affords  the  following 

EuLE.  Plot  the  equation.  Apply  the  location  principle  to  deter- 
mine between  what  consecutive  positive  integral  values  a  root  lies. 

Decrease  the  roots  of  the  equation  hy  the  lesser  of  the  two 
integral  values  between  which  the  root  lies. 

Estinfiate  from  the  plot  the  nearest  tenth  to  which  the  root  of 
the  new  equation  lies,  and  determine  hy  synthetic  division  pre- 
cisely the  successive  tenths  between  which  the  root  lies. 

Decrease  the  roots  of  this  equation  by  the  lesser  of  the  two  tenths 
between  ivhich  the  root  lies,  and  estimate  the  root  to  the  nearest 
hundredth  by  solving  the  last  two  terms  as  a  linear  equation. 


198 


ADVANCED  ALGEBRA 


Determine  hy  synthetic  division  precisely  the  hundredths  inter- 
val in  which  the  root  must  lie.  ^ — ^ 

Proceed  similarly  to  find  the  root  correct  to  as  many  'places  ds 
may  he  desired. 

The  sum  of  the  integral,  tenths,  and  hundredths  values  next  less 
than  the  root  in  the  various  processes  is  the  approximate  value 
of  the  root. 

To  find  the  negative  roots  of  an  equation  f(x)  =  0,  find  the 
positive  roots  of  /(—  x)  =  0  and  change  their  signs. 

When  all  the  roots  are  real  a  check  to  the  accuracy  of  the  com- 
putation may  be  found  by  adding  the  roots  together.  The  result 
should  be  the  coefficient  of  the  second  term. 


EXERCISES 

Eind  the  values  of  the  real  roots  of  the  following  equations  correct  to 
three  decimal  places. 

1,  x3  +  4x2  +  aj  +  l  =  0.    (1) 

Solution :  Since  by  Descartes' 
rule  of  signs  there  are  no  posi- 
tive roots,  we  form  the  equation 
/( —  a;)  =  0  and  seek  its  positive 
root. 

Thus 
a;3-4x2  +  aj-l  =  0.     (2) 

Plot  the  equation  (2)  set  equal 
to  y.  In  the  figure  two  squares 
are  taken  as  the  unit  of  x. 

There  is  a  root  of  this  equa- 
tion between  3  and  4. 


Yj 

< 

( 

X 

0 

i      ^ 

/ 

\ 

1 

> 

2 

k 

,f 

3 

\ 

4 

V 

J 

\ 

/ 

^ 

> 

y 
-1 

-3 

-7 
-7 
+  3 


Decrease  the  roots  of  (2)  by  3, 


1-4+1 
+  3-3 


1[3 
6 


1-1  -2 

+  3  +6 


1  +  2 

+  3 

1  +  5 


+  4 


THEORY  OF  EQUATIONS  199 

The  equation  is  x^  +  5x^  +  Ax  —  7  =  0.  (3) 

From  the  plot  we  estimate  the  root  of  (3)  at  .8. 

Verify,  1  +  5.0  +  4.00  -  7.000  [^ 

+  0.8  +  4.64  +  6.912 
+  5.8  +  8.64  -  0.088 

1+  5.0 +  4.00- 7.000  [^ 
+  0.9  +  5.31  +  8.379 
+  5.9+9.31  +  1.379 

Thus  the  root  is  determined  between  .8  and  .9.         \ 
Decrease  the  roots  of  (3)  by  .8, 

1+5.0  +    4.00  -7.000^8 
+  0.8  +    4.64  +  6.912 


1+5.8  +    8.64 
+  0.8  +    5.28 


.088 


1+6.6 
+  0.8 


+  13.92 


1+7.4 


The  equation  is        x^  +  7.4  x^  +  13.92  x  -  .088  =  0.  (4) 

088 

Estimate  the  root  of  (4)  at  x  =  ~ =  .006. 

^  '  13.92 

Verify,  1  +  7.400  +  13.920000  -  .088000 [.006 

+  0.006  +  00.044436  +  .083784 
+  7.406  +  13.964        -  .004216 
1  +  7 .400  +  13.920000  -  .088000 [.007 
+0 .007  +  00.051849  +  .097804 
+  7.407  +  13.972        +  .009804 
Thus  the  root  of  the  equation  (1)  correct  to  three  decimal  places  is  -  3.80(> 
2.  x3  -  4  =  0.  3.  x*  -  3  =  0. 

4.    X3  +  X  =  20.  5.    3X4  -  5X3  zz:  3I. 

6.  x3  +  x2  =  100.  7.  x3  -  X  -  33  =  0. 

8.  x4  +  X  -  100  =  0.  9.  x3  -  8x  -  24  =  0. 

10.  x4  -  4x3  +  12  =  0.  11.  X*  +  x2  +  X  =  111. 

12.  x3  -  x2  +  X  -  44  =  0.  13.  x3  +  lOx  -  13  =  0. 

14.  x8  +  3  x2  -  2  X  -  1  =  0.  15.  x3  +  x2  +  X  -  99  =  0. 

16.  x3  -  9x2  -  2 X  +  101  =  0.  17.  X*  +  x3  +  x2  -  88  =  0. 

18.  X*  -  12x3  -  16x  +  41  =  0.  19.  x3  -  6x2  +  5x  +  11  =  q. 

20.  x3  -  10x2 +  35x+ 50  =  0.  21.  2x4-4x3  +  3x2-1  =  0. 

22.  3x4  -  2x3  -  21x2 -4x  + 11  =  0.     23.  9x3  -  45x2  +  34x  +  37  =  0. 


200 


ADVANCED  ALGEBRA 


184.  Roots  nearly  equal.    Suppose  we  wish  to  find  the  positije 
roots,  if  any  exist,  of  ) 

x^-{-17x^-A6x-i-29  =  0.  (1) 


y 

4-29 

+  1 

+  13 
+  71 


By  Descartes*  rule  of  signs  we  see  that  there  can  be 
only  two  positive  roots.  We  obtain  the  adjacent  table 
of  values.  From  the  plot  that  these  values  indicate  we 
cannot  tell  whether  any  real  root  exists  between  1  and  2, 
but  if  it  does  exist  the  plot  indicates  that  it  is  nearer 
1  than  2. 

Decrease  the  roots  of  (1)  by  1, 

1  +  17  -46  +29[1 
+    1+18-28 


1  +  18  -  28 

+    1 

+    1+19 

1  +  19 

-    9 

+    1 

1+20 

The  new  equation  is 

x^-\-20x^- 

-9x- 

f  1  = 

0. 


(2) 


Estimate  the  root  of  (2)  at  .2  and  carry  the  origin  up  to  .2 


and  also  up  to  .3. 

1  +  20.0  -  9.00  +  l.OOOj^ 
+    0.2  +  4.04  -  0.992 


1  4-  20.0  -  9.00  +  1.000[J 
+    0.3  +  6.09  -  0.873 


1  +  20.2  -4.96 
+    0.2  +  4.08 


1  +  20.4 
+    0.2 


+  0.008 


1  +  20.3  -  2.91 
+    0.3  +  6.18 


0.88 


1  +  20.6 
+    0.3 


+  0.127 


+  3.27 


1  +  20.6 


1  +  20.9 


By  Descartes'  rule  of  signs  on  the  numbers  obtained  by  moving 
the  origin  to  .3,  it  is  seen  that  there  are  no  positive  roots  of  (2) 
greater  than  .3,  while  the  rule  would  indicate  that  there  might 
be  roots  greater  than  .2.    We  consider  the  equation 

ic^  +  20.6  x^  -  0.88  X  +  0.008  =  0.  (3) 


THEORY  OF  EQUATIONS  201 


Estimate  the  roots  of  (3)  at 
_  0.008 
^  ~  0.88 


=  .009.=* 


Verify,  1  +  20.60  -  0.880  -h  0.00800[.01 

+  0.01  +  0.206-0.00674 
1  -f  20.61  -  0.674  +  0.00126 
1  +  20.60  -  0.880  +  0.00800|.02 

+    0.02  +  0.412  -  0.00936 
1  +  20.62  -  0.468  -  0.00136 
This  determines  a  root  of  (3)  between  .01  and  .02. 

1  +  20.60  -  0.880  4-  0.00800|.03 

+  0.03  +  0.619-0.00483 
1  +  20.63  -  0.161  +  0.00317 

This  determines  another  root  of  (3)  between  .02  and  .03. 
Decrease  the  roots  of  (3)  by  .01, 

1  +  20.60  -  0.880  +  0.00800|.01 
+    0.01  +  0.206  -  0.00674 


1  +  20.61  -  0.674 
+    0.01  +0.206 


+  0.00126 


1  +  20.62 
+    0.01 


-  0.468 


1  +  20.63 

The  new  equation  is 

x^  +  20.63  x^  -  0.468  x  +  0.00126  =  0.  (4) 

Estimate  the  root  of  (4)  at  ic  =    '   ,^^    =  .002.=* 
^  ^  0.468 

Verify,  1  +  20.630  -  0.468  +  0.00126|.002 

+  0.002  +  0.041  -  0.00085 
1  +  20.632  -  0.427  +  0.00041 
1  +  20.630  -  0.468  +  0.00126|.003 

+  0.003  +  0.062-0.00122 
1  +  20.633  -  0.406  +  0.00004 

♦  We  observe  that  in  these  two  cases  the  estimated  values  of  the  roots  are  shown  by 
the  verification  to  be  inaccurate.  This  should  insure  great  care  in  making  the  verifi' 
cation.   Th^  estimated  values  should  nev^  b^  assumed  to  be  accurate  without  verification 


202  ADVANCED  ALGEBRA 

This  indicates  that  a  root  is  between  .003  and  .004. 

Verify,  1  +  20.630  -  0.468  +  0.00126|.004 

+    0.004  +  0.083-0.00154 
1  +  20.634  -  0.385  -  0.00028 

This  determines  a  root  of  (3)  between  .003  and  .004. 

Thus  one  root  of  (1)  correct  to  three  decimal  places  is  1.213. 

The  other  root  could  be  found  similarly  to  be  1.229. 

EXERCISES 

Find  all  the  real  roots  of  the  following  equations  correct  to  three  decimal 
places. 

1.  x^-1x-^1  =  0. 

2.  7x3 -8«2_i4x  + 16  =  0. 

3.  2ic5_  4x3-3x2  +  6  =  0. 

4.  4x4-  5x3 -8x  + 10  =  0. 

5.  3x3  -  10x2  -  33x  +  110  =  0. 


CHAPTEE   XVIII 
DETERMINANTS 

185.  Solution  of  two  linear  equations.  We  have  already 
treated  the  solution  of  linear  equations  in  two  variables  and 
stated  (p.  47)  the  method  of  solving  three  or  more  linear  equa- 
tions in  three  or  more  variables.  This  latter  process  is  rather 
laborious  and  can  be  very  much  abridged  and  also  developed 
more  symmetrically  by  the  considerations  of  the  present  chapter. 
Let  us  solve  the  equations 

a^x  +  b^ij  =  Ci,  (1) 

a^x  +  b^y  =  Ca.  (2) 

Multiply  (1)  by  b^  and  (2)  by  b^,  and  we  obtain 

ai^2^  4-  bib^i/  =  b2Ci  ^.     v 

Subtracting,  we  get  (aib^  —  a^bi)  x  =  b^c^  —  biC^, 

or  if  aA  -  ^2^1  ^  0,  ^^h^LIzh^.  (3) 

ai/>2  —  ^2^1 

Similarly,  we  obtain  y  =  —^ ^'  (4) 

We  note  that  the  denominators  of  the  expressions  for  x  and  y 
are  the  same.  This  denominator  we  will  denote  symbolically  by 
the  following  notation :  ^, 

aibz  —  a^bi  =  ,  •      _ 

^2        t*2 

The  symbol  in  the  right-hand  member  is  called  a  determinant. 
Since  there  are  two  rows  and  two  columns,  this  determinant  is 
said  to  be  of  the  second  order.  The  left-hand  member  of  the  equa- 
tion is  called  the  development  of  the  determinant.  The  symbols 
Ui,  bi,  a2,  ^2  ^^6  called  elements  of  the  determinant,  while  the  ele- 
ments «!  and  &2  are  said  to  comprise  its  principal  diagonal. 

203 


(6). 


204 


ADVANCED  ALGEBRA 


EuLE.    The  development  of  any  determinant  of  the  seconc 
order  is  obtained  by  subtracting  from  the  product  of  the  ele^ 
ments  on  the  principal  diagonal  the  product  of  the  elements  on 
the  other  diagonal. 


Thus 


Zi 


xyi  -  ziy ; 


12    3 
4    5 


10  -  12  =  -  2. 


Evidently  each  term  of  the  development  contains  one  and  only 
one  element  of  each  row  and  each  column,  that  is,  for  instance, 
the  letter  a  and  the  subscript  1  appear  in  each  term  of  (5)  once 
and  only  once. 

We  can  now  rewrite  the  solution  (3)  and  (4)  of  equations  (1) 
and  (2)  in  determinant  form : 


X  = 


Cl 

h 

ai     Ci 

C2 

h 

;  y  = 

^2        ^2 

ai 

b. 

a^     bi 

a^ 

b. 

a^     b^ 

(6) 


It  is  noted  that  the  numerator  of  the  expression  for  x  is  formed 
from  the  denominator  by  replacing  the  column  which  contains 
the  coefficients  of  x,  a^  and  a^,,  by  the  constant  terms  c^  and  c^. 
Similarly,  in  the  numerator  of  the  expression  for  y,  b^  and  b^  of 
the  denominator  are  replaced  by  Cj  and  c^. 

One  should  keep  in  mind  that  a  determinant  is  merely  a  sym- 
bolic form  of  expression  for  its  development.  In  the  case  of 
determinants  of  the  second  order  the  introduction  of  the  new 
notation  is  hardly  necessary,  as  the  development  itself  is  simple ; 
just  as  we  should  scarcely  need  to  introduce  the  exponential 
notation  if  we  had  to  consider  only  the  squares  of  numbers.  It 
turns  out,  however,  as  we  shall  see,  that  we  are  able  to  denote  by 
determinants  with  more  than  two  rows  and  columns  expressions 
with  whose  development  it  would  be  very  laborious  to  deal. 

186.  Solution  of  three  linear  equations.  Let  us  solve  the 
equations 


a^x  +  b^y  4-  Ci«  =  c^i, 
a^x  4-  b^2.y  +  ^2^  =  ^2> 
<iz^  +  b^y  +  Cg^  =  c?8- 


(1) 

(2) 
(3) 


DETERMINANTS 


205 


Eliminating  y  from  (1)  and  (2)  and  (1)  and  (3),  we  obtain 
(ai^g  —  a^h^x  H- (b^c^  —  hiCc^)z  =  dj)^  —  d^b^j 
(a^bi  —  a^b^x  +  {c^b^  —  b^c^z  —  d^b^  —  d-J)^. 

Eliminating  z, 

[iSHh  —  «2^i)  (^3^1  —  h(^i)  —  (ctsbi  —  a^bs)  (b^Ci  —  ^iC2)]a; 

=  (dibi  —  d^bi)  (c^bi  —  bsCi)  —  (d^bi  —  dj)^)  (b^Ci  —  b^c^} 

Developing,  canceling,  and  dividing  by  bi,  we  obtain 

_  d^b^Cs  +  d^bsCi  +  ^3^1  ^2  —  dib^c^  —  dsb^c^  —  d^b^c^ 

aib^c^  +  a^b^c^  +  a^b^c^  —  a^b^c^  —  a^b^c^  —  a^Jy^c^  ^ 

Following  the  analogy  of  tbe  last  section,  we  write  the  denomi- 
nator 


ai^2^3  +  «2^8^1  +  «8^1^2  ~"  %^3<'2  "~  «3^2Cl  ~  «2^1^3  = 


«1 

l\ 

Cl 

^2 

K 

^2 

^3 

bs 

Cs 

(5) 


The  right-hand  member  of  this  equation  we  call  a  determinant 
of  the  third  order,  and  the  left-hand  member  its  development. 
As  in  the  determinant  of  the  second  order,  the  elements  a^,  b^,  Cg 
comprise  the  principal  diagonal;  each  term  of  the  development 
contains  one  and  only  one  element  of  each  row  and  each  column, 
and  all  possible  terms  so  constructed  are  included  in  the  develop- 
ment. The  signs  of  the  terms  of  the  determinant  of  the  third 
order  may  be  kept  in  mind  by  the  following  device. 
'  Rewrite  the  first  and  second  columns  to  the  right  of  the  deter- 
minant as  follows : 


«!     ^1     G^ 

<Zi     &i 

%     ^2     ^2 

^2     h 

«3     ^8     ^3 

%    h 

The  positive  terms  are  found  on  the  diagonals  running  down 
from  left  to  right,  the  negative  terms  on  the  diagonals  running 
up  from  left  to  right. 

The  numerator  of  the  fraction  (4)  expressed  in  determinant 
notation  is 

«!        bi       Cj 

dz     b^     Cg 
dn     b^     Cg 


206 


ADVANCED  ALGEBRA 


We  can  find  similarly  the  values  of  y  and  z  that  satisfy  equ£ 
tions  (1),  (2),  and  (3).    The  solutions  of  the  equations  in  determi- 
nant form  are  as  follows  : 


d. 

^1 

^2 

d. 

b. 

d. 

h 

^3 

a^ 

K 

^1 

^2 

h 

<^2 

^3 

h 

Oz 

y  = 


^1 

d. 

Ci 

^2 

d. 

C2 

^3 

d. 

Cz 

«1 

h 

^1 

^2 

h 

^2 

^3 

h 

Cz 

«i 

*i 

d. 

aj 

*s 

d. 

as 

Js 

d> 

a, 

h 

h 

a. 

h 

<'2 

as 

h 

Cs 

(6) 


The  same  principle  observed  on  p.  204  for  forming  the  deter- 
minants in  the  numerators  of  the  expressions  for  x  and  y  may  be 
followed  here.  The  determinant  in  the  numerator  of  the  expres- 
sion for  ic,  2/,  or  z  is  found  from  the  denominator  by  replacing  the 
column  that  contains  the  coefficients  of  the  variable  in  question 
by  a  column  consisting  of  constant  terms.  Thus  in  the  numerator 
of  z  we  find  the  column  d-^,  d^^  d^  replacing  the  column  Cj,  Cg,  Cg  of 
the  denominator. 


EXERCISES 


1.  Find  the  value  of  the  following  determinants. 


3     2     1 

(a) 

4     6    2 
1     0     1 

• 

3     2     1 

Solution : 

4    6    2 

1     0     1 

4 

3 

1 

(b) 

1 

2 

0 

6 

1 

1 

3 

4 

1 

(d) 

0 

2 

4 

1 

0 

6 

a    X    y 

(f) 

0     b     c 
0     c     b 

• 

0       c     b 

(h) 

-c      0     a 

-b  -a 

o| 

=  18 +  4  +  0-6-8-0  =  8. 


2 

0 

1 

(c) 

2 

1 

0 

1 

2 

0 

2 

1 

1 

(e) 

6 

3 

3 

4 

2 

3 

a 

b 

c 

(g) 

b 

c 

a 

c 

a 

b 

0 

c 

b 

(i) 

c 

0 

a 

b 

a 

0 

DETERMINANTS 


207 


2.  Solve  the  following  equations  by  determinants. 


(a) 


2x  +  32/  =  4, 


x-2y  =  l. 
Solution :  Using  the  expressions  (6),  p.  206,  we  obtain 

-8 -3  _  n 
-4-3~y 


4 

3 

1 

-2 

2 

3 

1 

-2 

y  = 


2 

4 

1 

1 

2 

3 

1 

-2 

(e) 


7  y  =  12, 


7  a: +  2/ =  11. 


2x+5y=l,  2x  +  72/:-l,  a:  +  4y  =  2, 

^  '  7a;+6?/  =  2.    ^  -*  Sx-9y  =  2.    ^   '  2a;-3?/  = 

3.  Solve  the  following  equations  by  determinants. 

a:  +  y  +  2  =  2, 
(a)  X +  32/- 4  =  0, 

y  -  2  2;  =  6. 

Solution :  Rearranging  so  that  terms  in  the  same  variable  are  in  a  column, 
and  supplying  the  zero  coefficients,  we  get 

x+     2/4-     2  =  2, 

a:  +  32/  +  02  =  4, 

Ox+     y-2z  =  6. 


By  (6),  p.  206,  x 


2 

1 

1| 

4 

3 

0 

6 

1 

-2 

J 

1 

1 

1 

3 

0 

0 

1 

-2 

-12+0  +  4-18  +  8  +  0 


18 


1  I   /-^6  +  0  +  l 


0-0.+  2 


=  6, 


1 

2 

1 

1 

4 

0 

0 

6 

-2 

-8+0+6-0+4+0 

1 

1 

1 

-3 

1 

3 

0 

0 

1 

-2 

1  2 
3  4 
1       6 


1  1  1 
13  0 
0     1-2 

Check:  6  +  (-|)  +  (--L0)  = 


,18  +  0  +  2-0-6-4 
-3 


6-4  =  2. 


10 


10 


208 


\    AD 


2ic  +  3y  =  12, 
(b)  3a; +  2:2  =  1 

Sy  +  4:Z 
x  +  y  -z 
(d)  x  +  z~ 
V  +  z- 

x  +  y  + 
(f)  3a:-2z  = 
5  y  —  4  z  =  0. 

X  +  2/  +  2  =  9, 
(h)  x  +  2y  +  3z  =  14, 

jc  +  3y  +  62;  =  20. 

.2x  +  .3y +  .4z  =  29, 
(j)  .3x  +  .4y  +  . 521=38, 

.4a;  +  .5y  +  .6z  =  51. 


ED  AMe^EBRA 

7    ^lx-iy  =  0, 
Ac)  ix-iz  =  l, 
2;  -  i  y  =  2. 
a  +  2y  =  7, 
)  7x+92;  =  29, 
y+8z  =  17. 
X  +  y  +  2  2;  =  34, 
(g)  X  +  2  y  +  2:  =  33, 
2  X  +  y  +  2:  =  32. 
ax  +  6y  —  C2;  =  2  a6, 
(i)  6y  +  C2;  —  ax  =  2  6c, 
C2;  +  ax  —  6y  =  2  ac. 
3x  +  2y +  32:  =  110, 
(k)  5  X  +  y  -  4  2;  =  0, 
2x-3y  +  2;  =  0. 


187.  Inversion.  In  order  to  find  the  development  of  determi- 
nants with  more  than  three  rows  and  columns,  the  idea  of  an  inver- 
sion is  necessary.  If  in  a  series  of  positive  integers  a  greater 
integer  precedes  a  less,  there  is  said  to  be  an  inversion.  Thus  in 
the  series  12  3  4  there  is  no  inversion,  but  in  the  series  12  4  3 
there  is  one  inversion,  since  4  precedes  3.  In  1  4  2  3  there  are  two 
inversions,  as  4  precedes  both  2  and  3 ;  while  in  1  4  3  2  there  are 
three  inversions,  since  4  precedes  2  and  3,  and  also  3  precedes  2. 

188.  Development  of  the  determinant.  In  the  development  of 
the  determinant  of  order  three  we  have 

%^2^3  +  «2^3Cl  +  ^S^l^a  —  0^8^2Cl  —  ^2^1^8  "  «1^3^2-     (1) 

the  order  of  lettejps  in  each  term  the  same  as  their 
in  the  principal  diagonal  (as  we  have  done  in  the  development 
above),  it  is  observed  that  the  subscripts  in  the  various  terms 
take  on  all  possible  permutations  of  the  three  digits  1,  2,  and  3. 
The  permutations  that  occur  in  the  positive  terms  are  12  3, 
2  3  1,  3  1  2,  in  which  occur  respectively  0,  2,  and  2  inversions. 
The  permutations  that  occur  in  the  subscripts  of  the  negative 
terms  are  3  2  1,  2  1  3,  1  3  2,  in  which  occur  respectively  3,  1, 
and  1  inversions. 


DETERMINANTS  209 

Thus  in  the  subscripts  of  the  positive  terms  an  even  number 
of  inversions  occur,  while  in  the  subscripts  of  the  negative  terms 
an  odd  number  of  inversions  occur.  This  means  of  determining 
the  sign  of  a  term  of  the  development  we  shall  assume  in  general. 

When  we  have  a  determinant  with  71  rows  and  columns  it  is 
called  a  determinant  of  the  nth  order.  The  development  of  such  a 
determinant  is  defined  by  the  following 

KuLE.  The  development  of  a  determinant  of  the  nth  order  is 
equal  to  the  algebraic  sum  of  the  terms  consisting  of  letters  fol- 
lowing each  other  in  the  same  order  in  which  they  are  found  in 
the  principal  diagonal  hut  in  which  the  subscripts  take  on  all 
possible  permutations.  A  term  has  the  positive  or  the  negative 
sign  according  as  there  is  an  even  or  an  odd  number  of  inver- 
sions in  the  subscripts. 

This  means  of  finding  the  development  of  a  determinant  is  use- 
ful in  practice  only  when  the  elements  of  the  determinant  are 
letters  with  subscripts  such  as  in  (2)  below.  When  the  elements 
are  numbers  we  shall  find  the  value  of  the  determinant  by  a  more 
convenient  method. 

In  this  statement  it  is  assumed  that  the  number  of  inversions 
in  the  subscripts  of  the  principal  diagonal  is  zero.  If  this  number 
of  inversions  is  not  zero,  the  sign  of  any  term  is  +  or  —  accord- 
ing as  the  number  of  inversions  in  its  subscripts  differs  from  the 
number  in  the  subscripts  of  the  principal  diagonal  by  an  even  or 
an  odd  number. 

Since  each  term  contains  every  letter  a,  b,  ••-jk  and  also  every 
index  1,  2,  •  •  • ,  ri,  one  element  of  each  row  and  column  occurs  in 
each  term. 

In  the  determinant  of  the  fourth  order 

tti    61    Ci  di 

a^    62    C2  dz 

03    &3    C3  ^i 

a^    &4    C4  d^ 

the  terms  a^h^Cidi  and  a^h^c^di,  for  instance,  have  the  minus  sign,  as  2  4  1  3  haa 
three  inversions  and  4  2  3  1  has  five  inversions;  while  the  terms  aih^c^d^  and 
a^b^Cidi  have  two  and  six  inversions  respectively  and  hence  have  the  positive  sign. 


210 


ADVANCED  ALGEBRA 


189.  Number  of  terms.  We  apply  the  theorem  of  permutar 
tions  to  prove  the  following 

Theorem.  A  determinant  of  the  nth  order  has  n  !  terms  in 
its  development 

Since  the  number  of  terms  is  the  same  as  the  number  of  per- 
mutations of  the  n  indices  taken  all  at  a  time,  the  theorem  follows 
immediately  from  the  corollary  on  p.  145. 

190.  Development  by  minors.  In  the  development  of  the 
determinant  of  order  three,  p.  208,  we  may  combine  the  terms  as 
follows : 


<^l  (pi^^Z  —  h<^2)  —  »2  (P\<^Z  —  h(^l)  +  0^3  (^1^2  —  h<^i) 


3        Cs 


+  Cli 


(1) 


IS 


We  observe  that  the  coefficient  of  a^  is  the  determinant  that 
we  obtain  by  erasing  the  row  and  column  in  which  aj  lies.  A 
similar  fact  holds  for  the  coefficients  of  a^  and  a^.  The  determi- 
nant obtained  by  erasing  the  row  and  column  in  which  a  given 

element  lies  is  called  the  minor  of  that  element.    Thus    / 

the  minor  of  cig-  ^^  notice  that  in  the  above  development  by 
minors  (1)  the  sign  of  a  given  term  is  +  or  —  according  as  the 
sum  of  the  number  of  the  row  and  the  number  of  the  column 
of  the  element  in  that  term  is  even  or  odd.  Thus  in  the  first 
term  a^  is  in  the  first  row  and  the  first  column,  and  since  1  -f- 1  =  2, 
the  statement  just  made  is  verified  for  that  case.  Similarly,  aj  is 
in  the  first  column  and  the  second  row,  and  since  1  +  2  =  3  is 
odd,  the  sign  is  minus  and  the  law  holds  here.  The  last  term 
is  positive,  which  we  should  expect  since  a^  is  in  the  first  column 
and  the  third  row,  and  1  +  3  =  4.  The  proof  for  the  general 
validity  of  this  law  of  signs  is  found  on  p.  215. 

The  elements  of  any  other  row  or  column  than  the  first  may 
be  taken  and  the  development  given  in  terms  of  the  minors  with 


DETERMINANTS 


211 


—  (^2 

Cl 

+  ^'2 

«1 

Cl 

—  ^2 

a^ 

^3 

dz 

<^Z 

as 

respect  to  such,  elements.    For  instance,  take  the  development 
with  respect  to  the  elements  of  the  second  row, 

d^        ^2        ^2 

The  rule  of  signs  is  the  same  as  given  above ;  that  is,  for 
instance,  the  last  term  is  negative,  as  c^  is  in  the  third  column 
and  the  second  row,  and  2  +  3  =  5.  By  generalizing  these  con- 
siderations we  may  find  the  development  of  a  determinant  by 
minors  by  the  following 

EuLE.  Write  in  succession  the  elements  of  any  row  or  column, 
each  multiplied  by  its  minor. 

Give  each  term  a  +  or  a  —  sign  according  as  the  sum  of  the 
number  of  the  row  and  the  number  of  the  column  of  the  element 
in  that  term  is  even  or  odd. 

Develop  the  determinant  in  each  term  by  a  similar  process 
until  the  value  of  the  development  can  be  determined  directly  by 
multiplication. 

That  this  rule  for  development  gives  the  same  result  as  the  definition  given  in 
§  188  we  have  seen  for  a  determinant  of  order  three.  The  fact  holds  in  general,  as 
we  shall  prove  (p.  215) . 

EXERCISES 

1.  In  the  determinant  of  order  four  on  p.  209  what  sign  should  be  pre- 
fixed to  the  following  terms  ? 

(a)  Ct^hzO^di. 

Solution:  c^^hza-idi  =  azhcidi.  In  2  3  4  1  there  are  three  inversions.  The 
sign  should  be  minus. 

(b)  a^hcsdi.  (c)  aibic^ds.  (d)  b4Cidsa2- 
(e)  ^8610402.                        (f)  d2aic^b3.  (g)  0208^164- 

.   2.  Develop  by  minors  the  following  and  find  the  value  of  the  determinant 
2    4 


(a) 


Solution : 


1     4 

1     6 

3 

2 

3 


=  3 
=  3.(6 


4) 


|1 
2(12 


4)  +  3  (8  -  4)  =  6  -  16  +  12  =  2. 


212 


ADVANCED  ALGEBRA 


1    4    6 

(p) 

7     8    2 
1    3     1 
0    a    b 

(e) 

d    0     c 
e    f    0 

• 

2 

1 

0 

(c) 

2 

0 

1 

0 

3 

4 

0 

a 

b 

(f) 

a 

0 

b 

a 

6 

0 

(d) 

5    7 

3    7 

-2    3 

2 

1 
-  1 

(g) 

a    0    6 
0     c    0 
d    0     e 

• 

(h) 


Solution :  Develop  with  respect  to  the  elements  of  the  first  column, 


2  3 

3  \l 

4  2 
3     1 


112 
2  12 
12    3 

21 

2   31 

2 

3 


3  1  1 
2  12 
12     3 

1    2 


+  4 


3  11 
112 
12     3 


2    3 

1  1 

2  3 


+  1 


+  1 


:  s!)-('i; 


-2 


1 

2  3 

1  1 

1  2 


1  1 

1  2 

1  2 

1 


+  1 


+  2 


1    2 


(i) 


2    6    3 

i. ' 


=  2(-l  +  2  +  0)-3(-3-2  +  l)  +  4(-3-l  +  l)-3(0-l+2) 
=  2  +  12  -  12  -  3  =  - 1. 

1     2 


0  8 
0  6 
0    2 


9 
1  3 
1  8 
1     4 


ax 


1  1 
0  3 
0    3 


3  4 

1  1 

1  1 

3  3 


Hint.  It  is  always  advisable  to  develop  with  respect  to  the  row  or  column  that 
has  a  maximum  number  of  elements  equal  to  zero. 


(k) 


(m) 


(0) 


(q) 


a 

0 

0 

b 

b 

a 

0 

0 

0 

b 

a 

0 

0 

0 

b 

a 

2 

3 

4 

1 

2 

3 

3 

6 

2 

3 

1 

1 

2 

3 

2 

1 

X 

a 

b 

b 

X 

a 

. 

a 

b 

X 

X 

0 

0 

0 

y 

X 

0 

0 

0 

y 

X 

0 

0 

0 

y 

X 

0 

0 

0 

y 

(1) 


(n) 


(P) 


(r) 


«!     6i      Ci     di 

0     62    C2    di 


ai  0 
a2  62 
as 


C3  f^a 

0  ^4 

0  0 

0  0 


63 
a^    64    C4 
a    0    gr 


e 

a 

/ 

0 

0 

a 

d 

c 

b 

X 

a 

b 

c 

X 

a 

b 

c 

X 

0, 

6 

c 

cs     0 

0 
0 
0 
a 
c 
b 
a 

X 


DETERMINANTS 


213 


191.  Multiplication  by  a  constant.  In  this  and  the  following 
sections  we  shall  give  a  number  of  theorems  on  determinants 
which  greatly  facilitate  their  evaluation  and  which  make  a  proof 
for  the  solution  in  terms  of  determinants  of  any  number  of  linear 
equations  in  the  same  number  of  variables  a  simple  matter. 

Theorem.  If  every  element  of  a  row  or  a  column  is  multi- 
plied hy  a  number  m,  the  determinant  is  multiplied  by  m. 

Suppose  that  every  element  of  the  first  row  of  a  determinant 
is  multiplied  by  m.  Since  each  term  of  the  development  contains 
one  and  only  one  element  from  the  first  row,  every  term  is  multi- 
plied by  m,  that  is,  the  determinant  is  multiplied  by  m. 

Illustration. 

mbi     bi     ^3 
mci      Cg      C3 
=  maiJg^a  +  fnazb^Ci  -}-  ma^biC^  —  maib^c^,  —  ma^biC^  —  ma^b^Cx 


«!     a 

1     a. 

=  m 

bx      b^      ba 

Ci        C2       C3 

• 

6     4     1 

Similarly, 

8     3     2 

= 

10 

4     1 

23 
24 
2-5 


=  2 


192.  Interchange  of  rows  and  columns.   We  now  prove  the 

Theorem.    The  value  of  a  determinant  is  not  changed  if  the 
columns  and  rows  are  interchanged. 

Take  for  instance  the  determinant  of  order  four. 


«! 

bi 

Cl 

di 

ai 

Oa 

as 

a^ 

aa 

b. 

C2 

d. 

bi 

b. 

bs 

b. 

ttg 

bs 

Cz 

d. 

Ol 

^2 

C3 

Ci 

a^ 

b. 

C4 

d. 

d. 

d. 

d. 

d. 

In  each  of  these  determinants  the  principal  diagonal  is  the  same, 
and  hence  the  developments  derived  according  to  the  statement  on 


214 


ADVAl^CED  ALGEBRA 


p.  209  will  be  the  same  for  each  determinant,  since  the  terms  wil 
be  identical  except  for  their  order.  The  same  reasoning  is  valic 
for  any  determinant. 

193.  Interchange  of  rows  or  columns.    We  now  prove  the 

Theorem.    If  two  columns  or  two  rows  are  interchanged,  the 
sign  of  the  determinant  is  changed. 

Again  let  us  take  for  example  the  determinant  of  order  four  and 
fix  our  attention  on  the  first  and  second  rows.    We  must  prove  that 


«! 

h 

Ci 

d. 

a^ 

h. 

^2 

d. 

as 

h 

Cl 

d. 

- 

«! 

W 

Cl 

d. 

as 

h 

Cs 

ds 

as 

h 

Gz 

ds 

a^ 

h 

Ci 

d. 

a^ 

b. 

C4 

d. 

In  the  first  determinant  the  principal  diagonal  is  a^bc^c^d^,  while 
in  the  second  the  principal  diagonal,  a^b^^c^d^,  is  obtained  from  the 
principal  diagonal  of  the  first  determinant  by  one  inversion  of 
subscripts.  Hence  this  term  is  found  among  the  negative  terms 
of  the  first  determinant. 

Since  the  only  difference  between  the  second  determinant  and 
the  first  is  the  interchange  of  the  subscripts  1  and  2,  evidently 
any  term  of  the  second  is  obtained  from  some  term  of  the  first 
by  a  single  inversion.  Thus  if  a  single  inversion  is  carried  out  in 
every  term  of  the  first  determinant,  we  obtain  the  various  terms 
of  the  second.  But  since  this  process  changes  the  sign  of  each 
term  of  the  first  determinant  (p.  213),  we  see  that  the  second 
determinant  equals  the  negative  of  the  first.  Similar  reasoning 
may  be  applied  to  the  interchange  of  any  two  consecutive  rows 
or  columns  of  any  determinant. 

Consider  now  the  effect  of  interchanging  any  two  rows  which 
are  separated  we  will  say  by  k  intermediate  rows.  To  bring  the 
lower  of  the  two  rows  in  question  to  a  position  next  below  the 
upper  one  by  successive  interchanges  of  adjacent  rows,  we  must 
make  k  such  interchanges.  To  bring  similarly  the  upper  of  the  two 
rows  to  the  position  previously  occupied  by  the  other  requires  k-\-l 
further  interchanges  of  adjacent  rows.  Hence  the  interchange  of 
the  two  rows  is  equivalent  to  2  A;  + 1  interchanges  of  adjacent 


DETERMINANTS  215 

rows,  the  effect  of  which  is  to  change  the  sign  of  the  determinant, 
since  2  ^  +  1  is  always  an  odd  number. 

194.  Identical  rows  or  columns.    This  leads  to  the  important 

Theokem.  If  a  deteimiinant  has  two  rows  or  two  columns 
identical^  its  value  is  zero. 

Suppose  that  the  first  and  the  second  row  of  a  determinant  are 
identical.  Suppose  that  the  value  of  the  determinant  is  the  num- 
ber D.  By  §  193,  if  we  interchange  the  first  and  second  rows  the 
value  of  the  resulting  determinant  is  —  D.  But  since  an  inter- 
change of  two  identical  rows  does  not  change  the  determinant  at 
all,  we  have  j)  ___  t) 

or  2  Z>  =  0,  that  is,  D  =  0. 

Corollary.  If  any  row  (or  column)  is  m  times  any  other 
row  (or  column),  the  value  of  the  determinant  is  zero. 

By  §  191,  the  determinant  may  be  considered  as  the  product  of 
m  and  a  determinant  which  has  two  rows  (or  columns)  identical. 
Hence  this  product  equals  zero. 

195.  Proof  for  development  by  minors.  On  referring  to  the 
rule  on  p.  211  we  observe  that  in  order  to  show  that  the  develop- 
ment by  minors  is  the  same  as  the  development  obtained  by  the 
definition  on  p.  209  we  must  prove  the  two  following  statements. 

First.  The  coefficient  of  any  element  in  the  development  of  a 
determinant  (apart  from  sign)  is  the  minor  of  that  element. 

Second.  Each  element  times  its  minor  should  have  a  -\-  or 
a  —  sign  according  as  the  sum  of  the  number  of  the  row  and  the 
column  of  the  element  is  even  or  odd. 

Consider  the  element  ^j. 

First.  Each  term  that  contains  ai  must  contain  every  other 
letter  than  a,  and  the  indices  of  these  letters  must  take  on  all 
permutations  of  the  numbers  2,  3,  ■  •  -,  n.  This  coefficient  of  ai 
contains  then  by  definition  (p.  210)  all  the  terms  of  its  minor. 

Second.  Since  in  each  term  ai  is  in  the  first  place,  the  only  inver- 
sions in  the  subscripts  are  those  among  the  numbers  2,  3,  •  •  • ,  ?i. 


216 


ADVANCED  ALGEBRA 


Hence  the  sign  of  each  term  in  the  coefficient  of  ai  is  positive  or 
negative  according  as  there  is  an  even  or  odd  number  of  inversions 
in  its  subscripts.  Hence  our  theorem  is  established  for  the  ele- 
ment %. 

Consider  now  any  element,  as  d^ ,  which  occurs  in  the  fifth  row 
and  the  fourth  column.  Interchange  adjacent  rows  and  columns 
until  c?5  is  brought  into  the  leading  position  in  the  principal 
diagonal.  This  requires  in  all  seven  interchanges,  three  to  get 
the  c?6  in  the  first  column,  and  then  four  to  get  it  into  the  first 
row.  This  changes  the  sign  of  the  determinant  seven  times, 
leaving  it  the  negative  of  its  original  value.  By  the  reasoning 
just  given  in  the  case  of  %  the  coefficient  of  d^  (which  is  now  in 
the  position  previously  occupied  by  ai)  in  the  original  determi- 
nant would  be  the  minor  of  d^,  except  that  the  signs  would 
all  be  changed.  Hence  the  term  consisting  of  d^  times  its  minor 
has  the  —  sign,  and  the  theorem  is  proved  for  this  case. 

In  general,  to  bring  a  term  in  the  ith.  row  and  kth.  column  to 
the  leading  position  requires  i  —  l-{-k  —  l  =  i-\-k  —  2  inter- 
changes of  adjacent  rows  or  columns.  This  involves  i  -\-  k  —  2, 
or  what  amounts  to  the  same  thing,  i  +  k  changes  of  sign.  Hence 
a  positive  or  a  negative  sign  should  be  given  to  an  element  times 
its  minor  according  as  i  +  ^  is  even  or  odd. 

196.  Sum  of  determinants.  We  now  prove  the  fact  that  under 
certain  conditions  the  sum  of  two  determinants  may  be  written  in 
determinant  form.  The  fact  that  the  product  of  two  determinants 
is  always  a  determinant  is  extremely  important  for  certain  more 
advanced  topics  in  mathematics,  but  the  proof  lies  beyond  the 
scope  of  this  chapter. 

Theorem.  If  each  of  the  elements  of  any  row  or  any  column  con- 
sists of  the  sum  of  two  numbers,  the  determinant  may  he  written 
as  the  sum  of  two  determinants. 

For  example,  we  have  to  prove  that 


«!  +  a'l 

^1 

Cl 

«! 

h 

Cl 

a\ 

^1 

Cl 

a^  +  a\ 

h 

C2 

= 

^2 

h 

^2 

+ 

a'. 

h. 

Ci 

«»  -f  a's 

h 

^3 

^8 

h 

Cz 

«'s 

K 

Cs 

DETERMINANTS 


217 


Develop  the  first  determinant  by  minors  with,  respect  to  the 
first  column,  where  we  symbolize  the  minors  of  Ui  +  a'j,  a^  +  a'2, 
ftg  +  a's  by  Ai,  A^^  A^  respectively. 

We  have 


by  §  190, 


«!  +  a'l     bi     Ci 

^2  +  a'2        ^2        ^2 

as  +  a's      ^8      ^8 


«! 

^1 

Cl 

a\ 

h 

Cl 

a^ 

*2 

C2 

+ 

a\ 

h 

^2 

as 

^8 

Cs 

a\ 

h 

Cz 

=  (»!  -f  a'O^l  —  (^2  +  ^'2)^2  +(«8  +  a'8)^8, 

by  the  distributive  law, 

=  aj^i  —  a^Az  4-  0.3^3  +  a\Ay  —  a'2^2  +  «'8^3> 


by  §  190, 


The  method  of  proof  given  is  applicable  to  the  case  of  any  row 
or  column  of  a  determinant  of  order  n. 

197.  Vanishing  of  a  determinant.  For  the  solution  of  systems 
of  linear  equations  we  shall  make  use  of  the 

Theorem.  If  in  the  development  of  a  determinant  in  terms 
of  the  minors  with  respect  to  a  certain  column  (or  row)  the  ele- 
ments of  that  column  (or  row)  are  replaced  hy  the  elements  of 
another  column  (or  row),  the  resulting  development  equals  zero. 

For  example,  we  have 


«! 

h 

Ci 

d. 

a? 

h 

Ci 

d. 

as 

h 

Cz 

d. 

a^ 

h 

C4 

d. 

a-^A-i  —  a^A^  -p  ^sA^  —  a^A^j 


(1) 


where  an  A  represents  the  minor  of  the  a  with  the  same  sub- 
script. We  have  to  prove  that  if  we  replace  the  a's,  for  example, 
by  the  b'Sj  the  result  equals  zero,  that  is,  that 


biAi  —  b^A^  +  ^3^3  —  b^A^  —  0. 


(2) 


218 


ADVANCED  ALGEBRA 


This  expression  (2)  when  written  in  determinant  form  evidently 
would  have  the  same  form  as  the  left-hand  member  of  (1)  excej 
ing  that  the  first  column  would  consist  of  ^i ,  ^2  ?  ^3  j  ^4  •  ^^  should 
then  have  two  identical  columns  of  the  determinant,  which  would 
then  equal  zero  (§  194).  Thus  the  development  in  (2)  vanishes 
identically.    This  method  of  proof  is  perfectly  general. 

Corollary.  The  value  of  the  determinant  is  unchanged  if  the 
elements  of  any  row  (or  column)  are  replaced  hy  the  elements  of 
that  row  (or  column)  increased  hy  a  multiple  of  the  elements  of 
another  row  (or  column). 

Thus,  for  instance. 


«! 

^1 

Ci 

^2 

h. 

^2 

= 

as 

h 

Cs 

«! 

-i-nb. 

h 

Ci 

^2 

+  nh^ 

h 

^2 

ds 

+  nh. 

h 

Cz 

The  proof  follows  directly  from  §§  191,  196,  and  the  preceding 
theorem. 

198.  Evaluation  by  factoring.  If  in  a  determinant  whose 
elements  are  literal  two  rows  or  two  columns  become  identical 
on  replacing  a  by  h,  then  a  —  b  i^  a  factor  of  the  development. 
This  appears  immediately  from  §  160. 

Illustration.    Evaluate  by  factoring 


(1) 


Since  two  columns  become  identical  if  a  is  replaced  by  h,  a  by  c, 
or  h  by  c,  then  we  have  as  a  factor  of  the  development 

(a-b){b-c){c-a).  (2) 

To  determine  whether  the  signs  in  this  product  are  properly 
chosen,  that  is,  whether  the  development  should  contain  a  —  b  oy 
b  —  a,we  note  that  the  term  bc^  is  positive  in  the  development 
of  (1)  and  also  positive  in  the  expansion  of  (2).  Evidently  there 
is  no  factor  of  (1)  not  included  in  (2). 


1 

1 

1 

a 

b 

c 

a^ 

b' 

c^ 

DETERMINAKTS 


219 


199.  Practical   directions  for   evaluating   determinants.    In 

j&nding  the  value  of  a  numerical  determinant  the  object  is  to 
reduce  it  to  one  in  which  as  many  as  possible  of  the  elements 
of  some  row  or  column  are  zero.  One  should  ask  one's  self  the 
following  questions  on  attempting  to  evaluate  a  determinant: 

First.  Is  any  row  (or  column)  equal  to  any  other  row  (or 
column)  ?    If  so,  apply  §  191  for  the  case  m  =  0. 

Second.  Are  the  elements  of  any  row  (or  column)  multiples  of 
any  other  row  (or  column)  ?    If  so,  apply  §  191. 

Third.  If  we  add  (or  subtract)  a  multiple  of  the  elements  of 
one  row  (or  column)  to  the  elements  of  another,  will  an  element 
be  zero  ?    If  so,  apply  §  197. 

EXERCISES 

Evaluate  the  following  determinants. 


2 

3 

4 

3 

4 

5 

3 

2 

1 

2 

7 

6 

0 

1 

8 

7 

Solution :  We  obseive  that  if  we  subtract  each  element  of  the  first  column 
from  the  corresponding  element  of  the  second  column,  the  new  second  column 
has  every  element  1.  A  similar  result  is  obtained  by  subtracting  the  last  col- 
umn from  the  third  column.    Thus,  by  §  191, 


2    3 

4 

3 

2 

1 

1 

4    5 

3 

2 

4 

1 

1 

1     2 

7 

6 

"7 

1 

1 

1 

0     1 

8 

7 

0 

1 

1 

=  0. 


4  3     12 

6  113 

4  2     12 

3  6    2     1 


Solution :  Multiplying  the  last  column  by  2  and  the  whole  determinant  by  \ 
does  not  change  the  value  of  the  determinant  (§  191).    Thus 


4    3     12 

4    3     14 

6     113 

_  1 

6     116 

4     2     12 

~  2 

4     2     14 

3    6    2     1 

3     6     2     2 

220 


ADVANCED  ALGEBRA 


Subtracting  the  last  column  from  the  first  column  and  developing,  we 


obtam 


4 

3 

1 

4 

6 

1 

1 

6 

_  1 

4 

2 

1 

4 

~2 

3 

6 

2 

2 

0  3  14 
0  116 
0  2  14 
16    2    2 


3     14 

1  1    6 

2  14 


Subtracting  the  last  row  from  the  first  row, 


3     1 

1  1 

2  1 


0  0 

1  6 
1    4 


-,-(-2)  =  l. 


a  —  d  a  1 
b-d  b  1 
c  —  d     c     1 


a^  +  b 
2ab 


2ab 

1 

a2  +  62 

1 

2a6  +  62 

1 

5. 


7. 


9. 


3 

3 

4 

2 

1 

1 

2 

1 

2 

2 

3 

1 

2 

1 

3 

2 

0 

a; 

y 

2 

X 

0 

y 

2 

y 

2 

0 

« 

2 

y 

a; 

0 

a 

1 

0 

0 

& 

1 

1 

0 

c 

1 

2 

0 

d 

1 

3 

3 

10. 


4 

6 

8 

3 

1 

1 

2 

1 

2 

3 

4 

1 

2 

1 

3 

4 

110  1 
10  11 
0     111 

1111 


1 

1 

1 

0 

1 

1 

0 

1 

1 

0 

1 

1 

0 

1 

1 

1 

11. 


a 

a 

a 

a 

a 

b 

b 

b 

a 

b 

c 

c 

a 

b 

c 

d 

12. 


2 

1 

1 

1 

1 

2 

1 

1 

1 

1 

2 

1 

1 

1 

1 

2 

13. 


3 

7 

16 

14 

6 

15 

33 

29 

0 

1 

1 

1 

4 

2 

3 

1 

14. 


8 

2 

1 

4 

16 

29 

2 

14 

16 

19 

3 

17 

33 

39 

8 

38 

15. 


9 

13 

17 

4 

18 

28 

33 

8 

30 

40 

54 

13 

24 

37 

46 

11 

16. 


[2 

14 

16 

18 

2 

4 

6 

8 

4 

3 

2 

1 

3 

7 

11 

16 

DETERMINANTS 


221 


D, 


200.  Solution  of  linear  equations.  Suppose  that  we  have  given 
n  linear  equations  in  n  variables.  We  seek  a  solution  of  the  equa- 
tions in  terms  of  determinants.    For  simplicity,  let  n  —  4:.    Given 

a^x  +  hy  +  Ci«  +  d^w  =/i,  (1) 

a^x  -f  h^y  +  c^z  +  d^w  =/2,  (2) 

(^3^  +  hy  +  CsZ  +  d^w  =/8,  (3) 

a^x  +  hy  +  C4^  +  d^w  =/4.  (4) 

The  coefficients  of  the  variables  taken  in  the  order  in  which 
they  are  written^ift^  be  taken  as  forming  a  determinant  D,  which 
we  call  the  determinant  of  the  system.    Thus 
%     hi     Ci     di 

(^2        t>2        ^2        ftg 
^8        ^3        ^3        ^8 

a^     b^     C4     d^ 

Symbolize  by  ^1,  ^3,  etc.,  the  minors  of  %,  63,  etc.,  in  this  deter- 
minant. Let  us  solve  for  x.  Multiply  (1),  (2),  (3),  (4)  by  ^1,  A^, 
Azj  A^  respectively.    We  obtain 

A^a^x  +  A^biy  +  A^c^z  +  A^d^w  =  ^j/i, 
A^a^x  +  A^b^^y  +  ^3^2^  +  A^d^w  =  ^2/2, 
^gaga:  +  A^b^y  +  ^3^3^  +  ^af/gi^  =  ^3/3, 
A^a^x  -f-  ^4^'4?/  4-  A^c^z  +  ^4C?4i^;  =  A^f^. 

If  we  add  these  equations,  having  changed  the  signs  of  the 
second  and  fourth,  the  coefficient  of  x  is  the  determinant  Z>,  while 
the  coefficients  of  y,  z^  w  are  zero  (§  197).  The  right-hand  mem- 
ber of  the  equation  is  the  determinant  D,  excepting  that  the  ele- 
ments of  the  first  column  are  replaced  by  fufi,  fs,f^  respectively. 
Hence 


X  = 


/l 

h 

Cl 

d. 

A 

h 

Ca 

d. 

A 

h 

Cs 

d. 

A 

h 

^4 

d. 

«! 

h 

Cl 

d. 

^2 

h 

C2 

d. 

^8 

h 

Cz 

d. 

a^ 

h 

Ci 

d. 

222 


ADVANCED  ALGEBRA 


In  a  similar  manner  we  can  show  that  the  value  of  any  variable 
which  satisfies  the  equations  is  given  by  the  following 

EuLE.  The  value  of  one  of  the  variables  in  the  solution  of  n 
linear  equations  in  n  variables  consists  of  a  fraction  whose 
denominator  is  the  determinant  of  the  system  and  whose  numer- 
ator is  the  same  determinant,  excepting  that  the  column  which  con- 
tains the  coefficients  of  the  given  variable  is  replaced  by  a  column 
consisting  of  the  constant  terms. 

When  D  =  0,  we  cannot  solve  the  equations  unless  the  numer- 
ators in  the  expressions  for  the  solution  also  vanish. 
Illustration.    Solve  for  x 

ax  -\-2by  —  1, 

2by  +  ^cz  =  2, 

S  cz -{- 4:  dw  =  Sj 

4  dw  -{-  5ax  =  A. 

E-earranging,  we  obtain 

ax  +  2  by  =  ly 

2by-^3ez  =2, 

3  cz -{- 4:  dw  =  3j 

-{-  4:dW  =  4:. 


D 


5  ax 

a     2b  0 

0     2b  3c 

0      0  3c 

5a     0  0 


0 

v" 

0 

Ad 

=  24 

4:d 

a  5  0  0 

0  5  c  0 

0  0  c  ^ 

5a  ^  ^  d 


1 

b 

0 

0 

1 

b       0     0 

24 

2 

b 

c 

0 

1 

0       c     0 

3 

0 

c 

d 

3 

0       c     d 

X  — 

4 

0 

0 

d 

4 

0       0     d 

a 

b 

0 

0 

a 

b       0     0 

24 

0 

b 

c 

0 

0 

b       c     0 

0 

0 

c 

d 

0 

0      c     d 

5a 

0 

0 

d 

0 

-5b   0     d 

1 

c 

0 

-b 

3 

c 

d 

4 

0 

d 

b 

c 

0 

a 

0 

c 

d 

-5b 

0 

d 

DETERMINANTS 


223 


1 

c 

0 

-h 

2 

0 

d 

4 

0 

d 

b 

G 

0 

a 

0 

C 

d 

0 

5c 

d 

be 


ab 


G      d 
5c     d 


-  2  bed 
—  4  abed 


1_ 

2a 


201.  Solution  of  homogeneous  linear  equations.  The  equa- 
tions considered  in  the  previous  section  become  homogeneous 
(p.  115)  if  /i  =/2  =/3  =/,  =  0.    We  have  then 


a^x  +  b-^y  -\-  GiZ  +  diW  =  0, 
azX  -f  bzy  +  G2%  +  dzW  =  0, 
^3^5  +  b^y  +  ^3^;  +  cZgW;  =  0, 


(I) 


These  equations  have  evidently  the  solution  x  =  y  =  z  =  w  =0. 
This  we  call  the  zero  solution.  We  seek  the  condition  that  the 
coefficients  must  fulfill  in  order  that  other  solutions  also  may 
exist.  If  we  carry  out  the  method  of  the  previous  section,  we 
observe  that  the  determinant  equals  zero  in  the  numerator  of 
every  fraction  which  affords  the  value  of  one  of  the  variables 
(§  191).  Thus  if  D  is  not  equal  to  zero,  the  only  solution 
of  the  above  equations  is  the  zero  solution.  This  gives  us 
the  following 

Pkinciple.  a  system  of  n  linear  homogeneous  equations  in  n 
variables  has  a  solution  distinct  from  the  zero  solution  only 
when  the  determinant  of  the  system  vanishes. 

Whether  a  solution  distinct  from  the  zero  solution  always 
exists  when  the  determinant  of  the  system  equals  zero  we  shall 
not  determine,  as  a  complete  discussion  of  the  question  would  be 
beyond  the  scope  of  this  chapter. 

Theorem.  If  x^,  y^,  z^,  w^  is  a  solution  of  equations  (I) 
and  h  is  any  number,  then  kx^,  ky^,  kz^,  kw^  is  also  a  solution. 


224  ADVANCED  ALGEBRA 

The  proof  of  this  theorem  is  evident  on  substituting  kx^  etc., 
in  equations  (I)  and  observing  that  the  number  A;  is  a  factor  of 
each  equation.  Thus  if  a  system  of  n  linear  homogeneous  equar 
tions  has  any  solution  distinct  from  the  zero  solution  it  has  an 
infinite  number  of  solutions. 

EXERCISES 
Solve  for  all  the  variables : 

x  +  y  =  a,  a;  +  3  y  =  19, 

y  +  z  =  b,  y-]-Bz  =  S, 

M  +  «  =  d,  w  +  3d  =  11, 

V  +  x  =  e.  V  +  3ic  =  15. 

z-\-y  +  w  =  a,  Bx  +  y  +  z  =  20, 

^z  +  w  +  x  =  b,  -a;  +  4y  +  3w;  =  30, 

*w  +  a;  +  y  =  c,  '6jc  +  2  +  3iy  =  40, 

x-{-y-\-z  =  d.  8y  +  3z  + 6w;  =  50. 

2/  +  2;  +  5iy  =  ll,  x  —  2y  +  Sz  —  w  =  6, 

^    x  +  z-\-  4:W  =  llj  ^y-2z  +  Sw-x  =  0^ 

'a:  +  y  +  3M;  =  ll,  '   z-2w-\-Sx-y  =  0, 

x  +  z  +  Sy  =  3S.  w-2x  +  Sy  -  z  =  6. 


7. 


x  +  y-\-z  +  w  =  2i,  x  +  y  +  z  +  w  =  60, 

a;  +  2y  +  3z-9w  =  0,  g    x  +  2y  +  3z -\- ^w  =  100, 

3x-y  -6z-\-w  =  0,  '  x-\-Sy  -{■Qz-\-10w  =  150, 

2x  +  3y--4«- 610  =  0.  x  +  42/ +  lOz  +  20mj  =  210. 


CHAPTEE  XIX 
PARTIAL  FRACTIONS 

202.  Introduction.    For  various  purposes  it  is  convenient  to 

f(x) 
express  a  rational  algebraic  expression  ~j-\  ?  §  11,  as  the  sum  of 

several  fractions  called  partial  fractions,  which  have  the  several 
factors  of  <^{x)  as  denominators  and  which  have  constants  for 
numerators.  If  we  write  <j>(x)  =  {x  —  a)(x  —  P)---  (x  —  v),  we 
seek  a  means  of  determining  constants  Aj  B,  ■,  N  such  that  for 
every  value  of  x 

<l>{x)       X  —  a      X  —  p  X  —  V  ^  ^ 

If  the  degree  of  f(x)  is  equal  to  or  greater  than  that  of  ^  (a;), 
we  can  write 

<i>(x)  <}>(x)  ^  ^ 

where  Q  (x)  is  the  quotient  and  f^  (x)  the  remainder  from  dividing 
f(x)  by  <f>(x),  and  where  the  degree  oi  /^(x)  is  less  than  that  of 
<^  (x).  In  what  follows  we  shall  assume  that  the  degree  of  f(x) 
is  less  than  that  of  <f>  (x).  In  problems  where  this  is  not  the  case 
one  should  carry  out  the  long  division  indicated  by  (2)  and  apply 
the  principle  developed  in  this  chapter  to  the  expression  corre- 
sponding to  the  last  term  in  (2). 

203.  Development  when  </>(x)  =  0  has  no  multiple  roots.    Let 
us  consider  the  particular  case 

f(x)  ^  x  +  1 

<t>  (x)      (x  -l)(x-  2)  (x  -  3)' 

We  indicate  the  development  required  in  form  (1)  of  the  last 
section,  .  .  «  ^ 


(x-l)(x-2)(x-3)      x-1      x-2      x-S 
225 


226  ADVANCED  ALGEBRA 

where  A,  B,  and  C  represent  constants  which  we  are  to  deter- 
mine if  possible.  The  question  arises  immediately,  Are  we  at 
liberty  to  make  this  assumption?  Are  we  not  assuming  the 
essence  of  what'we  wish  to  prove,  i.e.  the  form  of  the  expansion? 
To  this  we  may  answer.  We  have  written  the  expansion  in  form  (1) 
tentatively.  We  have  not  proved  it  and  are  not  certain  of  its 
validity.  If,  however,  we  are  able  to  find  numerical  values  of 
Ay  B,  and  C  which  satisfy  (1),  we  can  then  write  down  the  actual 
development  of  the  fraction  in  the  form  of  an  identity. 

If,  on  the  other  hand,  we  can  show  that  no  such  numbers  A,BjC 
satisfying  (1)  exist,  then  the  development  is  not  possible. 

Clear  (1)  of  fractions, 

X  +  1  =  A(x  -  2){x  -  ^)+  B{x  -l){x  -  ^)-\-  C (x  -l)(x  -  2) 
=  {A  +  B-^C)x^-{6A  +4.B-\-^C)x^QA+^B-\-2C. 

Since  we  seek  values  of  A,  B,  and  C  for  which  (1)  is  identi- 
cally true  for  all  values  of  x,  equate  coefficients  of  like  powers  of 
X  in  the  last  equation  (Corollary  II,  p.  174).    We  obtain 

(2) 
(3) 
(4) 


(5) 
(2) 


A-\-B+  C  = 

-5A _45-3C= 

6A  +SB-\-2C  = 

=  0, 
-.1. 

Add  (4) 

to  (3)  and 

we  obtain 

Adding  we  obtain 

A -B- C= 

A+B-hC  = 

2A  = 

A  = 

=  2, 

:1     ^ 

From  (^] 

1  and  (4)  we  obtain 

B=-3j  C  = 

:2. 

Tlma 

x 

+  1 

1 

XIlUo 

(x-l)(x 

-2)(a.-3) 

X-1 

4- 


As  a  check  we  might  clear  of  fractions  and  simplify.  If  equar 
tions  (2),  (3),  and  (4)  had  been  incompatible,  we  should  have  con- 
cluded that  we  could  not  develop  the  fraction  in  form  (1). 


PARTIAL  FRACTIONS  227 

We  assume  now  for  the  general  case 

^(x)  =  (x  -  a)(x  -  P)--(x  -  v), 

and  that  the  n  roots  a,  )3,  •  •  •,  v  are  all  distinct  from  each  other. 
Let  us  consider  the  expression 

f(x)_      A  B  N 

<^{x)      X  —  a      X  —  p  X  —  V  ^ 

where  A,  B,  •",  N  are  constants.  Let  us  assume  for  the  moment 
the  possibility  of  expressing  ~y^  in  terms  of  these  partial  frac- 
tions. We  shall  now  attempt  to  determine  actual  values  A,  B,--, 
iV  which  satisfy  such  an  identity.  K  we  multiply  both  sides  of 
the  identity  by 

<t>(x)  =  (x-a)(x-P)-'(x-v), 
we  obtain 

f(x)  =  A(x-P)'"(x-v)  +  B(x-a)--'(x-v)  +  -" 

^N(x-a)(x-P)"; 

f(x)  is  of  degree  not  greater  than  n  —  1,  and  consequently  when 
written  in  the  form  of  (1),  p.  166,  has  not  more  than  n  terms.  If 
we  multiply  out  the  righ1>hand  member  and  collect  powers  of  ar, 
we  have  an  expression  in  a;  of  degree  n  —  1.  By  Corollary  II, 
p.  174,  this  equation  will  be  an  identity  if  we  can  determine 
values  ot  Aj  Bf  "-,N  which  make  the  coefficients  of  x  cm  both 
sides  of  the  equation  equal  to  each  other.  Hence  we  equate 
coefficients  of  like  jx>wers  of  x  and  obtain  n  equations  linear  in 
A,  B,  '",N  which  we  can  treat  as  variables.  These  equations  have 
in  general  one  and  only  one  solution  which  we  can  easily  deter- 
mine. The  values  of  Aj  B,  •  •  • ,  iV  obtained  by  solving  these  equa- 
tions we  can  substitute  for  the  numerators  of  the  partial  fractions 
in  (6).  After  making  this  substitution  we  can  actually  clear  of  • 
fractions  the  right-hand  member  of  (6)  and  check  our  work  by 
showing  its  identity  with  the  left-hand  member. 

There  is  no  general  criterion  that  we  have  applied  to  (6)  to 
determine  whether  the  n  linear  equations  obtained  lay  equating 
coefficients  of  x  have  any  solution  or  not.    Hence  in  this  general 


228  ADVANCED  ALGEBRA 

discussion  it  should  be  distinctly  understood  that  assumption  (6) 
holds  when  and  only  when  these  equations  are  solvable.  In 
any  particular  case  we  can  find  out  immediately  whether  the 
equations  are  solvable  by  attempting  to  solve  them.  If  the  num- 
bers Aj  By '  • ',  N  do  not  exist,  the  fact  will  appear  by  our  inability 
to  solve  the  linear  equations.  As  a  matter  of  fact,  one  and  only 
one  solution  always  exists  under  the  assumption  of  this  section. 

If  in  (6)  we  assume  that  several  of  the  symbols  A,  B,  ■'■,  N stand  for  expres- 
sions linear  in"  x,  as,  for  instance,  ax  +  b,  we  should  then  have  a  larger  number  of 
variables  to  determine  than  there  are  equations.  Under  these  circumstances  there 
is  an  infinite  number  of  solutions  of  the  equations.    Thus  if  we  should  seek  to 

express  =^-7—  as  the  sum  of  partial  fractions  where  the  numerators  are  not  con- 

stants  but  functions  of  z,  we  could  get  any  number  of  such  developments. 

We  have  the  following 

EtTLE.    Factor  <\>{x)  into  linear  factors,  as 
(x  —  a)(x  —  P)"-{x  —  v). 
Write  the  expression 

</)(a?)      X  —  a      X  —  P  x  —  v 

Multiply  both  sides  of  the  expression  by  <i>{x)y  equate  coefficients 
of  like  powers  of  x,  and  solve  the  resulting  linear  equations  for 
A,B,-^-,N. 

Replace  A,  B,  •",  N  hy  these  values  and  check  hy  substituting 
for  X  some  number  distinct  from  a,  ^,  "•,  v. 


EXERCISES 


Separate  into  partial  fractions ; 


(x-l)(x-2)x 


/J.2  _  2  A  B  f* 

Solution :  Assume = 1 1 — .  (1^ 

{x-l){x-2)x     a;-lx-2a;  ^' 

Multiply  by  (x  -  1)  (x  -  2)  x, 

x2  -  2  =  ^(x  -  2)x  +  J5(x  -  l)x  +  C(x  -  l)(x  -  2), 

3fi-2  =  {A+B-^C)x^-{2A+B  +  SC)z  -f  2^. 


PARTIAL  FKACTIOKS  229 


Equating  coefficients  of  like  powers  of  x, 
A  +  B  +  C  =  l, 
2A+B  +  SC  =  0, 
2  O  =  -  2. 
Hence  C  =  —  1. 

Solving 

we  obtain 


Thus 


A  +  Bz 

=  2, 

2A  +  B  = 

=  3, 

A  = 

=  1 

J5  = 

=  1 

■      '     +- 

1 

x^-2  1.1  1 


(a;-l)(x-2)a;      z-1      x-2      x 
Check :   Let 

Substituting  in  (1),        — -  =  — ~  -\ 


x  = 

-A, 

-1 

-6 

r 

-■2 

*^.- 

1_ 
6~ 

1 
2 

-b' 

5+1=1. 

6  -6 


2. 

x-1 

x^  -\-Sx-^2 

A 

1 

3  a;2  _  2  X  -  8 

(i 

4a;2 

{x^-i){x-l) 

R 

a;2_3a;  +  l 

3.         "  +  '' 


1. 


2a;2-5x-3 

5x 

6x2-5x-l 

2x2-1 

(x2  +  3a;  +  2)(x- 

-1) 

a;2  4-4 

9. 

(X  -  1)  (X  -  2)  (X  -  3)  (X  -  2)  (X  +  2)  (X  -  1) 

204.  Development  when  ^  (a?)  =  O  has  imaginary  roots.    In 

the  preceding  section  no  mention  has  been  made  of  any  distinc- 
tion between  real  and  imaginary  values  oi  a,  p,  ■  -- ,  v.  In  fact 
the  method  given  is  valid  whether  they  are  real  or  imaginary.  It 
is,  however,  desirable  to  obtain  a  development  in  which  only  real 
numbers  appear. 

Let  us  assume  the  development 

iS=-^  +  -^  +  -  +  ^^  +  -^'         (1) 

^  {x)      X  —  a      X  —  p  X  —  fji      X  —  V 

where  let  us  suppose  that  /x  and  v  are  the  only  pair  of  conjugate 
imaginary  roots  of  <^  (.r),  m  and  n  being  conjugate  complex  numbers. 


230  ADVANCED  ALGEBRA 

Let  fi  =  a-\-ib,  v  =  a  —  ib. 

Then  adding  the  corresponding  terms  of  (1),  we  obtain 

m  n        _  X  (m -{- n)  —  a  (m  +  n)  b(m  —  n) 

x  —  a  —  ib     x  —  a  +  ib  (x  —  a)^  -{-  b^  (x  —  af-\-b'^      ^  ^ 

Since  /a  and  y  are  the  only  imaginary  roots  of  <f>  (x)  =  0,  the  last 
term  of  (2)  is  real,  as  is  also  the  entire  right-hand  member  (§  152). 
Hence,  letting  the  numerator 

x(7}i-\-n)— a(m-\-n)-\-  ib  (m  —  n)  —  Mx  +  N, 

we  have  the  development 

/M^^4_       _B_  ■        Mx  +  N 

<^(x)      x-a      x-  ft  "^  (ic  _  a)2 -|.  J2  W 

By  complete  induction  we  can  establish  this  form  of  numerator 
where  there  is  any  number  of  pairs  of  imaginary  roots  of  <f>(x)  =  0. 
^e  have  proved  the  form  (3)  where  there  is  one  pair  of  imagi- 
nary roots.  Assuming  the  form  where  there  are  k  pairs,  we  can 
prove  it  similarly  where  there  are  A:  +  1  pairs.    Hence  we  have  the 

Theorem.  If  <l>(x)  is  facto7^ahle  into  distinct  linear  and  quad- 
ratic factors,  but  the  quadratic  factors  are  not  further  reducible 

fix) 
into  real  *  factors,  then  is  separable  into  partial  fractions 

of  the  form  ^^  ^ 

A            B  Mx  +  N 

+ ;;+•••+    o   ■  ■      > 


X  —  a      X  —  P  a?  -\-  ^x  +  V 

where  x^  +  fix  -\- v  is  an  irreducible  quadratic  factor  of  <t>(x). 

This  theorem  is  of  course  true  only  under  the  condition  that 
the  linear  equations  obtained  in  the  process  of  determining  the 
constants  are  solvable.  It  turns  out,  however,  that  in  this  case 
as  in  §  203  the  linear  equations  obtained  always  have  one  and 
only  one  solution  provided  that  the  roots  are  all  distinct. 

*  A  real /actor  is  one  whose  coefficients  are  all  real  numbers. 


PARTIAL  FRACTIONS 


231 


EXERCISES 


Separate  into  partial  fractions : 


x2+  1 


(X  -  1)  (x2  +  X  +  1) 
Solution :     Assi^me 


X2+1 


A  Bx  +  C 


...^        (X  -  1)  {X2  +  X  +  1)        X  -  1        X2  +  X  +  1 

Multiply  by  (x  -^{af^  +  a;  +  Ij,    - ' 

x2  +  1  =  ^  (x2  +  x  +  1)  +  (Bx  +  0)  («  -  1). 

Collecting  like  powers  of  x, 

3fi  +  l  =  {A  +  B)x^-{-  {A-  B+  C)x  +  A-C. 

Equating  coefficients  of  x, 

A  +  B  =  l, 

A-B-{-C  =  0, 

A-C  =  l. 

Add  (2)  and  (3)  and  solve  with  (1), 

2A-B=1 

A-{-B=:l 


Substituting  in  (1), 
Substituting  in  (3), 


SA  =  2, 

A  =  l 

B  =  h 


> 


(1) 
(2) 
(3) 


Thus 


X2  +  1 


x-1 


(x-l)(x2  +  x  +  l)      3(x-l)      3x2  +  3x  +  3 

Check:  Let  x  =  —  1. 

2  2  -2 


Substituting, 
Reducing, 

„     X2  +  a  +  1 


2-1      3 

6      3 


2      3-3+3 
1. 
x2  +  l 


x3  +  4x 

x2  +  4 

x8-2x2  +  3x- 

X 

2 

(X  +  3)  (2  x2  -  X 

X2  +  1 

-4) 

X*  +  X2  +  1 

5x^-1 

x4  +  6x2  +  8* 

1 


(X  -  l)(3x2  +  x  +  6) 


x3  +  3x2-2x-16 
Hint.    Factor   by    synthetic    divi- 
sion (see  p.  178) . 


232  ADVANCED  ALGEBRA 

205.  Development  when  ^  (x)  =  (x  —  a)~.  In  this  case  the 
method  given  in  the  previous  sections  fails,  as  the  equations  for 
determining  the  values  of  the  numerators  are  incompatible.    If 

^®^®*  f{x)  ^  a,x-'  +  a,x-'  +...  +  a^_^ 

<f>(x)  (x-ay  '  ^  ^ 

we  can  separate  into  partial  fractions  as  follows. 

Let  X  —  a  =  y,  that  is,  x  =  y  -\-  a,  and  substitute  in  (1).  We 
obtain  after  collecting  powers  of  y 

_ = 1 — -^ f_  — _, 

y  y     y  y 

where  the  ^'s  are  constants.    Eeplacing  yhyx  —  a,wQ  have  th^ 
following  development : 

A, 


/(f)       =      ^0       I  ^1  I  ^'^  I    ....). 

(x  —  a)"      X  —  a      (x  —  ay      (x  —  ay  (x  —  ay 


EXERCISES 

Separate  into  partial  fractions : 

-    x2  +  2  a;  -  1 

(X-3)8 

Solution :  Assume     — -—  = H — ■  -(- — -.  (1) 

(X  -  3)8  jc  -  3       (X  -  3)2       (X  -  3)8  ^  ' 

Multiply  by  (x  -  3)8, 

x2  4-  2 X  -  1  =  ^ (x2  -  6x  +  9)  +  5(x  -  3)  +  C 

Collecting  powers  of  x, 

=  Ax'^  +  {B-6A)x-\-9A-BB-\-C. 
Equating  coefficients  of  x, 

^  =  1, 

5-6^  =  2, 

9A-SB  +  C  =  -1. 

Solving,  B  =  8,  C  =  U. 

„                     x2  +  2x-l          1,8  14 

Hence  — ' = 1 \- 


(X  -  3)8  X  -  3        (X  -  3)2        (X  -  3)8 

Check :  Let  x  =  1. 

^  ^  ,.,,..     ,,,         2  1         8      14  1      „      14 

Substituting  in  (1),     __  =  — -  +  --_=_-  +  2-— , 

1  1 


PARTIAL  FRACTIONS  233 


2.  ^^^..  3. 


(x-2)3  (2a; +  1)2  (cc  -  4)3 

g    x^  +  x  +  1  g      a;-  g  -    2 a;^  +  3 x  +  1 

(2x-l)*"  '  (ax +  6)2*  *      (3x-2)3 

206,  General  case.  When  <f>(x)=  0  has  real,  complex,  and 
multiple  roots,  we  may  use  all  the  previous  methods  simultane- 
ously.   Hence  for  this  case  we  assume  the  expansion 

f(^ 

(x  —  a)--- (Xx^+  fxx  -\-  v) ■  " (x  —  tY 


X  —  a  Ax2-f  fjiX  -\-  V  X  —  T  (£c  —  t)' 


EXERCISES 

Separate  into  partial  functions  : 
-         g^  +  2x2  +  i8x-18 
'  (x-l)(x2  +  x  +  l)(x-2)2' 

Solution : 

x*  +  2x2  +  18x-18  A  Bx-VG  D      .        E 


(X  -  1)  (x2  +  X  +  1)  ^  -  2)2      X  -  1      x2  +  X  +  1      X  -  2      (X  -  2)2 

=  ^(x2  +  X  +  1)  (x  -  2)2  +  {Bx  +  C)  (X  -  2)2 (X  -  1) 
+  i)(x-l)(x-2)(x2  +  x  +  l) 

+  JSJ(X-1)(X2  +  X  +  1) 

=  {A-hB+D)x^-\-{-3A-6B  +  C-2D-\-E)x^ 
-{-{A  +  8B-5C)x2  +  (-4E  +  8C'-D)x 
+  (4  ^  -  4  C  +  2  D  -  ^). 

Equating  coefficients  of  like  powers  of  x, 

A+    B  +     D  =1,  (1) 

-SA-6B-\-     C-2D  +  E  =  0,  (2) 

A-^8B-5C  =2,  (3) 

-  4  B  +  8  C  -     D  =18,  (4) 

4  A  -4C  +  2D-^  =  -18.  (6) 

Adding  (2)  and  (5),  (1)  and  (4),  we  have,  together  with  (3), 

^_5E-3C'  =  -18,  (6) 

A-3B-\-SC  =  19,  (7) 

A-\-SB-6C  =  2.  (8) 


234  ADVANCED  ALGEBRA 

Adding  all  three  equations,  we  find 

3^  =  3,  or  A  =  l. 

Substituting  in  (3)  and  (7)  and  solving,  we  find  C  =  S,  B  =  2.    Substituting 
in  (1),  we  find  D  =  -  2.    Similarly,  from  (6),  ^  =  6. 

x*  +  2a;2  +  18x-18              1      ,     2a;  +  3           2      ,        6 
Thus — — -  = +  — 4- 


(x-l)(x2  +  a;  +  l)(x-2)2      x-1      x^  +  x  +  1      x-2      (a; -2)2 

Check:    Let  x  =  —  1. 

go  J         2         2        6 

Substituting,  -^-^  =  __  +  ____  +  _, 

¥  =  2i-i  =  -V-. 
2.  -^!±^.  3. 

X{X-  1)3 

.      X3  -f  2  X2  +  1  _ 

4. o. 

X(X-1)3 

g    2x3  +  x  +  3  y 

(X2  +  1)2 

8  ^~^  9 

■  (x  +  l)2(x  +  2)' 

10.        ^^^  +  ^      .  11. 

(a:2-l)(x  +  2) 

12.     ^^-^^  +  ^   .  13. 

(x  -  8)  (X  -  9) 

..          2x2 -3x- 3  .- 

14. 15. 


16.  "^^ — tl 17. 

(2x-3)(6x2-6x  +  l)  (a;2_3a;4.2)(x-3) 


l-x* 

5X  +  12 

x(x2  +  4) 

43X-11 

30(x2-l) 

X3  -  X  -  1 

X4-16 

x2  +  6x-8 

x3-4x 

2 

(x2  +  X  +  3)  (2  a 

•  +  1) 

x3  4-2x2-3x 

+  1 

(x  +  3)(x2-4x 

+  5) 

13 -5x 

CHAPTEE  XX 
LOGARITHMS 

207.  Generalized  powers.  If  h  and  c  are  integers,  we  can  easily 
compute  h''.  When  c  is  not  an  integer  but  a  fraction  we  can  com- 
pute the  value  of  ¥  to  any  desired  degree  of  accuracy.  Thus  if 
6  =  2,  c  =  I,  we  have  2^  =  V2*  =  VS,  which  we  can  find  to  any 
number  of  decimal  places.  If,  however,  the  exponent  is  an  irrar 
tional  number  as  V2,  we  have  shown  no  method  of  computing 
the  expression.  Since,  however,  V2  was  seen  (p.  ^5)  to  be  the 
limit  approached  by  the  sequence  of  numbers 

1,  1.4,  1.41,  1.414,  ••, 

it  turns  out  that  5^^  is  the  limit  approached  by  the  numbers 


The  computation  of  such  a  number  as  S^"*^  would  be  somewhat 
laborious,  but  could  be  performed,  since  5^-^^  =  5**^  =  V5^*\  Thus 
it  is  a  root  of  the  equation  x^^  =  S^'*^  and  could  be  found  by 
Horner's  method,  p.  197. 

We  see  in  this  particular  case  that  5^^  is  the  limit  approached 
by  a  sequence  of  numbers  where  the  exponents  are  the  successive 
approximations  to  V2  obtained  by  the  process  of  extracting  the 
square  root.  In  a  similar  manner  we  could  express  the  meaning 
of  5",  where  i  is  a  positive  integer  and  c  is  any  irrational  number. 

Assumption.  We  assume  that  the  laws  of  operation  which  we 
have  adopted  for  rational  exponents  hold  when  the  exponents  are 
irrational. 

Thus  6c.6d=  6c+d^  ^=  6e-d,   {Tbd)c=  {ho)d=  bcd^ 

where  c  and  d  are  any  numbers,  rational  or  irrational. 

236 


236  ADVANCED  ALGEBRA 

208.  Logarithms.  We  have  just  seen  that  when  h  and  c  are 
given  a  number  a  exists  such  that  ¥  =  a.  We  now  consider  the 
case  where  a  and  h  are  given  and  c  remains  to  be  found.  Let 
a  =  8,  ^  =  2.  Then  if  2*^  =  8,  we  see  immediately  that  c  =  3  satis- 
fies this  equation.  If  a  =  16,  5  =  2,  then  2"  =  16  and  c  =  4  is  the 
solution.  If  a  =  10,  6  =  2,  consider  the  equation  2"  =  10.  If  we 
let  c  =  3,  we  see  that  2^  =  8.  If  we  let  c  equal  the  next  larger 
integer,  4,  we  see  2*  =  16.  If  then  any  number  c  exists  such  that 
2*^  =  10,  it  must  evidently  lie  between  3  andj:.  To  prove  the  exist- 
ence of  such  a  number  is  beyond  the^^ope  of  this  chapter,  but 
we  make  the  following 

Assumption.  There  always  exists  a  real  number  x  which 
satisfies  the  equation 

b^  =  a,  (1) 

where  a  and  h  are  positive  numbers,  provided  b  ^  1. 

Since  any  real  number  is  expressible  approximately  in  terms  of 
a  decimal  fraction,  this  number  x  is  so  expressible. 

The  power  to  which  a  given  number  called  the  base  must  be 
raised  to  equal  a  second  number  is  called  the  logarithm  of  the 
second  number. 

In  (1)  X  is  the  logarithm  of  a  for  the  base  b. 

This  is  abbreviated  into 

X  =  \ogf,a.  (2) 

Thus  since 

28  =  8,  102  =  100,  3-  2  =  J,  40  =  1, 
we  have 

3  =  logaS,  2  =  logiolOO,  -  2  =  logg  j,  0  =  log4l. 

The  number  a  in  (1)  and  (2)  is  called  the  antilogarithm. 

EXERCISES 

1.  In  the  following  name  the  base,  the  logarithm,  and  the  antilogarithm, 
and  write  in  form  (2). 

(a)  36  =  729. 

Solution :   3  =  base,  6  =  logarithm,  729  =  antilogarithm,  logs  729  =  6. 

(b)  2*  =  16.  (c)  38  =  27. 


LOGARITHMS  237 

2.  Find  the  logarithms  of  the  following  numbers  for  the  base  3 : 

81,  243,  1,   i,   ij. 

3.  For  base  2  find  logarithms  of  8,  128,  |,  ^V 

4.  What  must  the  base  be  when  the  following  equations  are  true  ? 
(a)  log  49  =  2.  (b)  log81  =  4. 

(c)  log  225  =  2 .  (d)  log  625  =  4. 

209.  Operations  on  logarithms.  By  means  of  the  law  expressed 
in  the  Assumption,  §  207,  we  arrive  at  principles  that  have  made 
the  use  of  logarithms  the  most  helpful  aid  in  computations  that 
is  known. 

Theorem  I.  The  logarithm  of  the  prodicct  of  two  numbers  is 
the  sum  of  their  logarithms. 

Let  log^a.  =  a;, 

logfeC  =  y. 
Then  by  (1)  and  (2),  p.  236,    h^  =  a, 

¥  =  c. 
Multiply  (by  Assumption,  §  207), 

or  by  (1)  and  (2),  \og^a-c  =  x -\- y. 

Theorem  IL  The  logarithm  of  the  nth  power  of  a  number  is 
n  times  the  logarithm  of  the  number. 


Let 

logi,a  =  Xy 

or 

b'  =  a. 

Raise  both  sides  to  the  nth.  power, 

(b^f.  =  b'^  =  a% 

or 

logft  a"  =  nx. 

Example. 

Iogiol00  =  2, 

logio  1000  =  3. 

By  Theorem  I, 

logio  100,000  =  5, 

which  is  evidently  true, 

since              10^  =  100,000. 

238  ADVANCED  ALGEBRA 

Theorem  III.    The  logarithm  of  the  quotient  of  two 
is  the  difference  between  the  logarithms  of  the  numbers. 

Let  logj,  a  =  Xj 

log6C  =  2/. 
Then  b''  =  a, 

b^  =  c. 

Dividing,  ^,«-y  =  -, 

c  ^^ 

,       a 
or  log;,  -==x  —  y. 

G 

Theorem  IV.    The  logarithm  of  the  real  nth  root  of  a  nur.t- 
ber  is  the  logarithm  of  the  number  divided  by  n. 

Let  logj,  a  =  Xj 

or  b""  =  a. 

1  X 

Extract  the  nth  root,    (b'^y  =  ^^  =  -Va, 

or  logj  -Va  =  -  • 

n 

EXERCISES 

Given  logio2  =  .301,    logio5  =  .699,   logio?  =  .8451,  find 

1.  log(-s/f^-  V5).* 
Solution : 

By  Theorem  I,     log  (  v^T^  •  VS)  =  log  v^  +  log  V5. 
By  Theorems  III  and  IV,  =  f  log  7  +  ^  log  5. 

Now  f  log  7  =  f  (.  8451)  = .  50706, 

and  i  log  5  =  |_(.699)    =  .349^ 

Adding,  |  log  7  +  |  log  5  =  log  (  Vt^  .  VB)  =  .85656 

2.  log  40.  3.  log  28. 
Hint.   Let  40=  8-5=  28.6. 

4.  log  140.  5.  log  V280. 

6.  log  V'35.  7.  log  ( V8  .  v^ .  V^). 

8.  log  ( V5 .  78).  9.  log  ("t/Ie .  Vli .  4^700), 

•  Where  no  base  is  written  it  is  assumed  that  the  base  10  is  employed. 


LOGARITHMS  239 

210.  Common  system  of  logarithms.  For  purposes  of  compu- 
tation 10  is  taken  as  a  base,  and  unless  some  other  base  is  indi- 
cated we  shall  assume  that  such  is  the  case  for  the  rest  of  this 
chapter.  We  may  write  as  follows  the  equations  which  show  the 
numbers  of  which  integers  are  the  logarithms. 

Since  10^  =  100,000  we  have  log  100,000  =A 


10*  =  10,000 

log  10,000 

=  4. 

108  ^   IQQQ 

log  1000 

=  3. 

102  ^   IQQ 

log  100 

=  2. 

10^  =   10 

log  10 

=  1. 

10«  =  1 

logl 

=  0. 

io-»  =  .i 

log.l 

=  -1. 

10-2  =  .01 

log  .01 

=  -2. 

10- « =  .001 

log  .001 

=  -3. 

etc. 

etc, 

Assuming  that  as  x  becomes  greater  log  x  also  becomes  greater, 
we  see  that  a  number,  for  example,  between  10  and  100  has  a 
logarithm  between  1  and  2.  In  fact  the  logarithm  of  any  number 
not  an  exact  power  of  10  consists  of  a  whole-number  part  and  a 
decimal  part. 

Thus  since  IQS  <  3421  <  lO^, 

log  3421  =  3.  +  a  decimal. 
Since  10- 3  <  .0023  <  10-2, 

log  .0023  =  -  3.  +  a  decimal. 

The  whole-number  part  of  the  logarithm  of  a  number  is  called 
the  characteristic  of  the  logarithm. 

The  decimal  part  of  the  logarithm  of  a  number  is  called  the 
mantissa  of  the  logarithm. 

The  characteristic  of  the  logarithm  of  any  number  may  be  seen 
from  the  above  table,  from  which  the  following  rules  are  imme- 
diately deduced. 

'f  The  characteristic  of  the  logarithm  of  a  number  greater  than  unity 
is  one  less  than  the  number  of  digits  to  the  left  of  its  decimal  point. 

Thus  the  characteristic  of  the  logarithm  of  471  is  2,  since  471  is  between  100 
and  1000;  of  27.93  is  1,  since  this  number  is  between  10  and  100;  of  8964.2  is  3, 
since  this  number  is  between  1000  and  10,000. 


240  ADVANCED  ALGEBRA 

The  characteristic  of  the  logarithm  of  a  number  less  than  1 
is  one  greater  negatively  than  the  nwmher  of  zeros  ^preceding  the 
first  significant  figure. 

Thus  the  characteristic  of  the  logarithm  of  .04  is  —  2 ;  of  .006791  is  —  3 ;  of 
.4791  is  - 1. 

It  must  constantly  be  kept  in  mind  that  the  logarithm  of  a 
number  less  than  1  consists  of  a  negative  integer  as  a  character- 
istic plus  a  positive  mantissa.  To  avoid  complication  it  is  desir- 
able always  to  add  10  to  and  subtract  10  from  a  logarithm  when 
the  characteristic  is  negative.  Thus,  for  instance,  instead  of  writ- 
ing the  logarithm  -  3  +  .4672  we  write  10  -  3  +  -4672  -  10,  or 
7.4672  —  10.  This  is  convenient  when  for  example  we  wish  to 
divide  a  logarithm  by  2,  as  by  Theorem  IV,  §  209,  we  shall  wish 
to  do  when  we  extract  a  square  root.  Since  in  the  logarithm 
—  3  -f  .4672  the  mantissa  is  positive,  it  would  not  be  correct  to 
divide  —  3.4672  by  2,  as  we  should  confuse  the  positive  and 
negative  parts.  This  confusion  is  avoided  if  we  use  the  form 
7.4672  - 10,  and  the  result  of  division  by  2  is  3.7336  -  5,  or 
8.7336  —  10.  The  actual  logarithm  which  is  the  result  of  this 
division  is  —  2  +  .7336. 

Theorem.  Numhers  with  the  same  significant  figures  which 
differ  only  in  the  position  of  their  decimal  points,  have  the  same 
mantissa. 

Consider  for  example  the  numbers  24.31  and  2431. 

Let  10^  =  24.31. 

.    Then  a  =  log  24.31. 

If  we  multiply  both  numbers  of  this  equation  by  100,  we  have 
10^10^  =  10^+2^2431, 
or  x-\-2  =  log  2431. 

Thus  the  logarithm  of  one  number  differs  from  that  of  the  other 
merely  in  the  characteristic.  In  general  numbers  with  the  same  sig- 
nificant figures  are  identical  except  for  multiples  of  10.  Hence  their 
logarithms  differ  only  by  integers,  leaving  their  mantissas  the  same. 

Thus  if  log47120.  =  4.6732,  log47.12  =  1.6732,  and  log  .004712  =  - 3.6732,  or 
7.6732  -  10. 


LOGARITHMS  241 


EXERCISES 


By  §209, 

log  V600  =  1.38905 

2.  log  .06. 

3.  log  (210)3. 

4.  log 

5.  log(4.2)4. 

6.  log 

"^2.1 

7.  log 

a    ,      567 

Q    ,      13.23 

9.  log 

^  1.28 

If  log  2  =  .3010,  log  3  =  .4771,  log7  =  .8451,  find 
1.  logVOOO. 

Solution :        log  V600  =  log  V20  •  30  =  ^  log  20  +  i  log  30. 

By  the  preceding  theorem,        log 20  =  1.3010,  log 30  =  1.4771. 
i  log  20=    .6505 
I  log  30=    .73855 


(70)3 
324 


211.  Use  of  tables.  A  table  of  logarithms  contains  the  man- 
tissas of  the  logarithms  of  all  numbers  of  a  certain  number  of 
significant  figures.  The  table  found  later  in  this  chapter  gives 
immediately  the  mantissas  for  all  numbers  of  three  significant 
figures.  In  the  next  section  a  method  is  given  for  finding  the 
mantissa  for  a  number  of  four  figures.*  Hence  the  table  is  called 
a  four-place  table.  Before  every  mantissa  in  the  table  a  decimal 
point  is  assumed  to  stand,  but  in  order  to  save  space  it  is  not 
written.  To  find  the  logarithm  of  a  number  of  three  or  fewer 
significant  figures  we  apply  the  following 

Rule.    Determine  the  characteristic  hy  rules  in  §  210. 

Find  in  column  N  the  first  two  significant  figures  of  the  num- 
ber.   The  mantissa  required  is  in  the  row  with  these  figures. 

Find  at  the  top  of  the  page  the  last  figure  of  the  number. 
The  mantissa  required  is  in  the  column  with  this  figure. 

When  the  first  significant  figure  is  1  we  may  find  the  loga- 
rithm of  any  number  of  four  figures  by  this  rule  from  the  table 
on  pp.  248,  249  if  we  find  the  first  three  instead  of  the  first  two 
figures  in  column  N. 

Thus  the     log  516.  =  2.7126,        log  .00281  =  -  3.4487,        log  7400.  =  3.8692, 
log  600.  =  2.7782,  log  50.  =  1.6990,  log  4.00  =  .6021. 


242  ADVANCED  ALGEBRA 

EXERCISES 
Find  the  logarithms  of  the  following  : 

1.  3.  2.  303.  3.  .024. 

4.  347.  6.  .0333.  6.  1.011. 

7.  .202.  8.  .0029.  9.  .0001. 

10.  .00299.  11.  68400.  12.  .0201. 

212.  Interpolation.  We  find  by  the  preceding  rule  that, 
log  2440  =  3.3874,  while  log  2450  =  3.3892.  If  we  seek  the  loga- 
rithm of  a  number  between  2440  and  2450,  say  that  of  2445, 
it  would  clearly  be  between  3.3874  and  3.3892.  Since  2445  is  just 
halfway  between  2440  and  2450,  we  assume  that  its  logarithm  is 
halfway  between  the  two  logarithms.  To  find  log  2445,  then,  we 
look  up  log  2440  and  log  2450,  take  half  (or  .5)  their  difference, 
and  add  this  to  the  log  2440.    This  gives 

log  2445  =  3.3874  +  .5  x  .0018  =  3.3883. 
If  we  had  to  find  log  2442  we  should  take  not  half  the  difference  but 
two  tenths  of  the  difference  between  the  logarithms  of  2440  and 
2450,  since  2442  is  not  halfway  between  them  but  two  tenths  of  the 
way.  This  method  is  perfectly  general,  and  we  may  always  find  the 
logarithm  of  a  number  of  more  than  three  figures  by  the  following 

Rule.  Annex  to  the  proper  characteristic  the  mantissa  of 
the  first  three  significant  figures. 

Multiply  the  difference  between  this  mantissa  and  the  next 
larger  mantissa  in  the  table  (called  the  tabular  difference  and 
denoted  by  D)  by  the  remaining  figures  of  the  number  preceded 
by  a  decimal  point. 

Add  this  product  to  the  extreme  right  of  the  logarithm  of  the 

first  three  figures,  rejecting  all  decimal  places  beyond  the  fourth. 

In  this  process  of  interpolation  we  have  assumed  and  used  the  principle  that 
the  increase  of  the  logarithm  is  proportional  to  the  increase  of  the  number.  This 
principle  is  not  strictly  true,  though  for  numbers  whose  first  significant  figure 
is  greater  than  1  the  error  is  so  small  as  not  to  appear  in  the  fourth  place  of  the 
logarithm.  For  numbers  whose  first  significant  figure  is  less  than  2  this  error 
would  often  appear  if  we  found  the  fourth  place  by  interpolation.  For  this  reason 
the  table  on  pp.  248,  249  gives  the  logarithms  of  all  such  numbers  exact  to  four 
figures,  and  in  this  part  of  the  table  we  do  not  need  to  interpolate  at  all 


LOGARITHMS  243 


EXERCISES 

Find  the  logarithms  of  the  following : 
1.  63.924. 

Solution:  log  63.9  =  1.8055 

2 


Tabular  difference  = 


4.  62230. 

7.  20060. 

10.  9.999. 

13.  5.7828. 

16.  3.1416. 

log  275  =2.4393 

D=  16 

6 
2.4399 

.4 
6.4 

log63.924  =  1.8057  iL 

1.68 
We  add  2  to  1.8055  rather  than  1,  because  1.68  is  nearer  2  than  1.    In  general 
we  take  the  nearest  integer. 

2.  269.4.  3.  1001. 

5.  392.8.  6.  9.365. 

8.  .4283.  9.  .3101. 

11.  82.93.  12.  .05273. 

14.  .003011.  15.  .002156. 

17.  276.4  X  1.463. 
Solution :  log  275.4  =  2.4399 

log  1.463  =  0^652 
By  Theorem  II,  §  209,  log  (275.4  x  1.463)  =  2.6051 

18.  374.3  X  1396.  19.  1.46  x  237.2. 
20.  469.1  X  63.92.                               21.  47320.  x  .8994. 

22.  :5?!?1. 

^8-^^  log  .0372  =8.5705 -10    /)=  12 

Solution:      log. 03724  =  8.5710  -  10                                      5 _A 

log38.46    =L585^_                      ^.^^J^'^  ,M 

6.9860  -  10                                              7  .6 

1.5850  7^ 

23   5:^.  24   1^51??. 

.2364  '    5.128 

213.  Antilogarithms.  We  can  now  find  the  product  or  quotient 
of  two  numbers  if  we  are  able  to  find  the  number  that  corresponds 
to  a  given  logarithm. 

For  this  process  we  have  the  following 

KuLE.  If  the  mantissa  is  found  exactly  in  the  table,  the  first 
two  figures  of  the  corresponding  number  are  found  in  the  column 
N  of  the  same  row,  while  the  third  figure  of  the  number  is  found 
at  the  top  of  the  column  in  which  the  mantissa  is  found. 

Place  the  decimal  point  so  that  the  rules  in  §  210  are  fulfilled. 


244  ADVANCED  ALGEBRA 

EXERCISES 

Find  the  antilogarithms  of  the  following: 

1.  3.7419. 
Solution:  We  find  the  mantissa  .7419  in  the  row  which  has  55  in  coT 

umn  N.  The  column  in  which  .7419  is  found  has  2  at  the  top.  Thus  the 
significant  figures  of  the  antilogarithm  are  552.  Since  the  characteristic  is  3, 
we  must  by  the  rule  in  §  210  have  four  figures  to  the  left  of  the  point. 
Thus  the  number  sought  is  5520. 

2.  1.3874.  3.  2.7050.  4.  .6785. 
5.  2.8414.*                           6.  5.8831.  7.   1.5752. 
8.  9.9112  - 10.     ■              9.  3.7251.                          10.  6.3997. 

If  the  mantissa  of  the  given  logarithm  is  between  two  man- 
tissas in  the  table,  we  may  find  the  antilogarithm  by  the  following 

EuLE.  Write  the  number  of  three  figures  corresponding  to  the 
lesser  of  the  two  mantissas  between  which  is  the  given  mantissa. 

Subtract  this  mantissa  from  the  given  mantissa,  and  divide 
this  number  by  the  tabular  difference  to  one  decimal  place. 

Annex  this  figure  to  the  three  already  found,  and  place  the 
decimal  point  as  the  rules  in  §  210  require. 

It  should  be  kept  in  mind  that  we  may  always  add  and  subtract 
any  integer  to  a  logarithm.    This  is  useful  in  two  cases : 

First.  When  we  wish  to  subtract  a  larger  logarithm  from  a 
smaller ; 

Second.  When  we  wish  to  divide  a  logarithm  by  an  integer 
that  is  not  exactly  contained  in  the  characteristic. 

Both  these  processes  are  illustrated  in  exercise  2  (1)  following. 

EXERCISES 

1.  Find  the  antilogarithms  of  the  following : 

(a)  2.3469. 

Solution :  The  mantissa  3469  is  between  3464  and  3483.    Hence  D  =  19. 

The  mantissa  3464  corresponds  to  222.  To  find  the  fourth  significant  figure 
of  the  antilogarithm,  divide  3469  -  3464  =  5  by  D  =  19.  Since  5  -^  19  =  .26, 
we  annex  3  to  222.    Hence  the  antilogarithm  =  222.3. 

*  We  write  -  2  +  .8414  in  the  form  2.8414  to  save  space  and  at  the  same  time  to  recall 
the  fact  that  the  mantissa  is  positive. 


LOGARITHMS 


245 


(b)  4.3147.  (c)  1.5271.  (d)  1.4216. 

(e)  1.6423.  (f)  2.8791.  (g)  .7214. 

2.  Perform  the  following  operations  by  logarithms. 
1375  X  .06423 


(a) 


76420 


Solution : 

log  1375=    3.1383 
log. 06423=    8.8077-  10 

Adding  (Theorem  I, 

§  209),                                   11.9460  -  10 
log  76420=    4.8832 

Subtracting  (Theorem  III,  §  209),  log  result  =    7.0628  -  10 

result  =      .001156. 

(b)  (11)8. 

(c)  iUUY-                           (d)  5871  -, 

-  9308. 

(e)  (H)«. 

(f)  (3f I)*-".                           (g)  7066  -. 

-5401. 

(h)  8308  X  .0003769. 

(i)  3410  X  .008763. 

8.371  X  834.6 

,,  ,  37.42  X  11.21 

^■"^           7309 

^^^         BBAl        ■ 

^1)    8/87xV7194 
\    98080000 

Solution : 

log  87=    1.9395 

By  Theorem  IV,  §  209,      i  log  7194  =    1.9285 

Adding, 

=  13.8680  -  10 
log  98080000=    7.9916 

By  Theorem  III,  §  209,                       3)25.8764  -  30 

log  result  =    8.6255  -  10 

result  =      .04222. 

Since  in  the  subtraction  in  this  problem  we  have  to  subtract  7  from  3, 
we  add  and  subtract  10  to  the  minuend  to  avoid  a  negative  logarithm.  Since 
in  the  division  by  3  we  would  have  a  remainder  in  dividing  —  10  by  3,  we 
add  and  subtract  20  so  that  3  may  be  exactly  contained  in  30,  the  negative 
part  of  the  logarithm. 


(m)  ^. 

(n)  V:06. 

(0)   ^(.043)8. 

(P)^^. 

(q)  (.21)§. 

(r)  m  V'lOO. 

(s)  •^:o3. 

(t)   ^100. 

(u)   V(l. 563)3. 

(v)  V.00614. 

(w)  VHi- 

(X)  ^0.9  VI!- 

(y)  ■>J^-v^ 


(z)  >y.47  VM- 


214.  Cologarithms.    The  logarithm  of  the  reciprocal  of  a  num- 
ber is  called  its  cologarithm.    When  a  computation  is  to  be  made 


246 


ADVANCED  ALGEBRA 


in  which  several  numbers  occur  in  the  denominator  of  a  fraction, 
the  subtraction  of  logarithms  is  conveniently  avoided  by  the  use 
of  cologarithms.    By  our  definition  we  have 

colog  25  =  log  3jV  =  log  1  -  log  25,       Theorem  III,  §  209 
log  1  =  10.  -  10 

log  25  =    1.3979 
colog  25=    8.6021-10 

Thus  in  dividing  a  number  by  25  we  may  subtract  the  logarithm 
of  25,  or  what  amounts  to  the  same  thing,  add  the  logarithm  of 
^ij,  which  is  by  definition  the  cologarithm  of  26. 

EuLE.  The  cologarithm  of  any  number  is  found  by  subtract- 
ing its  logarithm  from  10  —  10. 

In  the  process  of  division  subtracting  the  logarithm  of  a  num- 
ber and  adding  its  cologarithm  are  equivalent  operations. 


EXERCISES 


Compute,  using  cologarithms. 
-     8  X  62.73  X  .052 


66  X  8.793 
Solution : 


2-  MlVlf- 
4.   V38.462  -  15.382. 
g    5086  (.0008769)8 
9802  (.001984)*  * 


</: 


'  'XVsox 


log  8=  .9031 

log  62. 73=  1.7975 

log.  052=  8.7160-10 

colog  56=  8.2518-10 

colog  8.793  =  9.0559  -  10 
log  result  =  27.7243 -30 

result  =  .00^99 


3.   V1572  -  872. 

Hint  .  1572  -  872  =  (157  +  87)  (157  -  87) 
=  (244)  (70). 


6.   V(27:5)2  -  (3.483)2. 


693  X  .04692 
03841  X  (569.8)2 


3  X  421.6  X 


.046.(200.1)^ 


(68.96)^  X  86.61 
.09263  V^ 

1416  X  (5.638)2 
(75)i 


LOGARITHMS  247 

215.  Change  of  base.  We  have  seen  that  the  logarithm  of  a 
number  for  the  base  10  may  be  found  to  four  decimal  places  in 
our  tables.  It  is  occasionally  necessary  to  find  the  logarithm  of  a 
number  for  a  base  different  from  10.  For  the  sake  of  generality, 
we  assume  that  the  logarithms  of  all  numbers  for  a  base  b  are 
computed.  We  seek  a  means  of  finding  the  logarithm  of  any 
number,  as  x,  for  the  base  c ;  that  is,  we  seek  to  express  log^a?  in 
terms  of  logarithms  for  the  base  h. 

Suppose  logc  X  =  z,  that  is,  c^  =  x. 

Take  the  logarithm  of  this  equation  for  the  base  h,  and  we  have 

logftC^  =  Z  log^C  =  logftOJ. 


Then 

^      log,a^ 

l0g6« 

If  we  let 

M  =  l0gf,Gj 

we  have 

..      ^og,x 

(1) 

This  number  M  does  not  depend  on  the  particular  number  x, 
but  only  on  the  two  bases.  From  (1)  we  see  that  we  can  find  the 
logarithm  of  any  number  for  the  base  c  by  dividing  its  logarithm 
for  the  base  b  by  M.  The  number  M  is  called  the  modulus  of  the 
new  system  with  respect  to  the  original  one. 

EuLE.  To  find  the  logarithm  of  a  member  for  a  new  base  c, 
divide  the  common  logarithm  by  the  modulus  of  the  system  whose 
base  is  c. 

EXERCISES 


Find: 

1.  logs  21. 

solution:    log321.>^-^;  =  lf^^f  =  2.771. 
logioo        .4771 

2.  log5  6.                              3.  logalS. 

4.    l0gi6  2. 

5.  logs  167.                           6.  logi8  237. 

7.  log2.i6l.41. 

248 


ADVANCED  ALGEBRA 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

100 

0000 

0004 

0009 

0013 

0017 

0022 

0026 

0030 

0035 

0039 

101 
102 
103 

104 

105 
106 

107 
108 
109 

0043 
0086 
0128 

0170 
0212 
0253 

0294 
0334 
0374 

0048 
0090 
0133 

0175 
0216 
0257 

0298 
0338 
0378 

0052 
0095 
0137 

0179 
0220 
0261 

0302 
0342 
0382 

0056 
0099 
0141 

0183 
0224 
0265 

0306 
0346 
0386 

0060 
0103 
0145 

0187 
0228 
0269 

0310 
0350 
0390 

0065 

0107. 

0149 

0191 
0233 
0273 

0314 
0354 
0394 

0069 
0111 
0154 

0195 
0237 
0278 

0318 
0358 
0398 

0073 
0116 
0158 

0199 
0241 
0282 

0322 
0362 
0402 

0077 
0120 
0162 

0204 
0245 
0286 

0326 
0366 
0406 

0082 
0124 
0166 

0208 
0249 
0290 

0330 
0370 
0410 

110 

0414 

0418 

0422 

0426 

0430 

0434 

0438 

0441 

0445 

0449 

111 
112 
113 

114 
115 
116 

117 

118 
119 

0453 
0492 
0531 

0569 
0607 
0645 

0682 
0719 
0755 

0457 
0496 
0535 

0573 
0611 
0648 

0686 
0722 
0759 

0461 
0500 
0538 

0577 
0615 
0652 

0689 
0726 
0763 

0465 
0504 
0542 

0580 
0618 
0656 

0693 
0730 
0766 

0469 
0508 
0546 

0584 
0622 
0660 

0697 
0734 
0770 

0473 
0512 
0550 

0588 
0626 
0663 

0700 
0737 
0774 

0477 
0515 
0554 

0592 
0630 
0667 

0704 
0741 
0777 

0481 
0519 
0558 

0596 
0633 
0671 

0708 
0745 
0781 

0484 
0523 
0561 

0599 
0637 
0674 

0711 
0748 

0785 

0488 
0527 
0565 

0603 
0641 
0678 

0715 
0752 
0788 

120 

0792 

0795 

0799 

0803 

0806 

0810 

0813 

0817 

0821 

0824 

121 
122 
123 

124 

125 
126 

127 

128 
129 

0828 
0864 
0899 

0934 
0969 
1004 

1038 
1072 
1106 

0831 
0867 
0903 

0938 
0973 
1007 

1041 
1075 
1109 

0835 
0871 
0906 

0941 
0976 
1011 

1045 
1079 
1113 

0839 
0874 
0910 

0945 
0980 
1014 

1048 
1082 
1116 

0842 
0878 
0913 

0948 
0983 
1017 

1052 
1086 
1119 

0846 
0881 
0917 

0952 
0986 
1021 

1055 
1089 
1123 

0849 
0885 
0920 

0955 
0990 
1024 

1059 
1093 
1126 

0853 
0888 
0924 

0959 
0993 
1028 

1062 
1096 
1129 

0856 
0892 
0927 

0962 
0997 
1031 

1065 
1099 
1133 

0860 
0896 
0931 

0966 
1000 
1035 

1069 
1103 
1136 

130 

1139 

1143 

1146 

1149 

1153 

1156 

1159 

1163 

1166 

1169 

131 
132 
133 

134 
135 
136 

137 

138 
139 

1173 
1206 
1239 

1271 
1303 
1335 

1367 
1399 
1430 

1176 
1209 
1242 

1274 
1307 
1339 

1370 
1402 
1433 

1179 
1212 
1245 

1278 
1310 
1342 

1374 
1405 
1436 

1183 
1216 
1248 

1281 
1313 
1345 

1377 
1408 
1440 

1186 
1219 
1252 

1284 
1316 
1348 

1380 
1411 

1443 

1189 
1222 
1255 

1287 
1319 
1351 

1383 
1414 
1446 

1193 
1225 
1258 

1290 
1323 
1355 

1386 
1418 
1449 

1196 
1229 
1261 

1294 
1326 
1358 

1389 
1421 
1452 

1199 
1232 
1265 

1297 
1329 
1361 

1392 
1424 
1455 

1202 
1235 
1268 

1300 
1332 
1364 

1396 
1427 
1458 

140 

1461 

1464 

1467 

1471 

1474 

1477 

1480 

1483 

1486 

1489 

141 
142 
143 

144 
145 
146 

147 
148 
149 

1492 
1523 
1553 

1584 
1614 
1644 

1673 
1703 
1732 

1495 
1526 
1556 

1587 
1617 
1647 

1676 
1706 
1735 

1498 
1529 
1559 

1590 
1620 
1649 

.  1679 
1708 
1738 

1501 
1532 
1562 

1593 
1623 
1652 

1682 
1711 
1741 

1504 
1535 
1565 

1596 
1626 
1655 

1685 
1714 
1744 

1508 
1538 
1569 

1599 
1629 
1658 

1688 
1717 
1746 

1511 
1541 
1572 

1602 
1632 
1661 

1691 
1720 
1749 

1514 
1544 
1575 

1605 
1635 
1664 

1694 
1723 
1752 

1517 
1547 

1578 

1608 
1638 
1667 

1697 
1726 
1755 

1520 
1550 
1581 

1611 
1641 
1670 

1700 
1729 
1758 

150 

1761 

1764 

1767 

1770 

1772 

1775 

1778 

1781 

1784 

1787 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

LOGARITHMS 


249 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

150 

1761 

1764 

1767 

1770 

1798 
1827 
1855 

1884 
1912 
1940 

1967 
1995 
2022 

1772 

1775 

1778 

1781 

1784 

1787 

151 
152 
153 

154 
155 
156 

157 

158 
159 

1790 
1818 
1847 

1875 
1903 
1931 

1959 
1987 
2014 

1793 
1821 
1850 

1878 
1906 
1934 

1962 
1989 
2017 

1796 
1824 
1853 

1881 
1909 
1937 

1965 
1992 
2019 

1801 
1830 
1858 

1886 
1915 
1942 

1970 
1998 
2025 

1804 
1833 
1861 

1889 
1917 
1945 

1973 
2000 
2028 

1807 
1836 
1864 

1892 
1920 
1948 

1976 
2003 
2030 

1810 
1838 
1867 

1895 
1923 
1951 

1978 
2006 
2033 

1813 
1841 
1870 

1898 
1926 
1953 

1981 
2009 
2036 

1816 
1844 
1872 

1901 
1928 
1956 

1984 
2011 
2038 

160 

2041 

2044 

2047 
2074 
2101 
2127 

2154 
2180 
2206 

2232 

2258 
2284 

2049 

2052 

2055 

2057 

2060 

2063 

2066 

161 
162 
163 

164 
165 
166 

167 
168 
160 

2068 
2095 
2122 

2148 
2175 
2201 

2227 
2253 
2279 

2071 
2098 
2125 

2151 
2177 
2204 

2230 
2256 
2281 

2076 
2103 
2130 

2156 
2183 
2209 

2235 
2261 

2287 

2079 
2106 
2133 

2159 
2185 
2212 

2238 
2263 
2289 

2082 
2109 
2135 

2162 
2188 
2214 

2240 
2266 
2292 

2084 
2111 
2138 

2164 
2191 
2217 

2243 
2269 
2294 

2087 
2114 
2140 

2167 
2193 
2219 

2245 
2271 

2297 

2090 
2117 
2143 

2170 
2196 
2222 

2248 
2274 
2299 

2092 
2119 
2146 

2172 
2198 
2225 

2251 
2276 
2302 

170 

2304 

2307 

2310 

2312 

2315 

2317 
2343 
2368 
2393 

2418 
2443 
2467 

2492 
2516 
2541 

2320 

2322 

2325 

2327 

171 
172 
173 

174 
175 
176 

177 

178 
179 

2330 
2355 
2380 

2405 
2430 
2455 

2480 
2504 
2529 

2333 
2358 
2383 

2408 
2433 
2458 

2482 
2507 
2551 

2335 
2360 
2385 

2410 
2435 
2460 

2485 
2509 
2533 

2338 
2363 
2388 

2413 
2438 
2463 

2487 
2512 
2536 

2340 
2365 
2390 

2415 
2440 
2465 

2490 
2514 

2538 

2345 
2370 
2395 

2420 
2445 
2470 

2494 
2519 
2543 

2348 
2373 
2398 

2423 
2448 
2472 

2497 
2521 
2545 

2350 
2375 
2400 

2425 
2450 
2475 

2499 
2524 
2548 

2353 
2378 
2403 

2428 
2453 
2477 

2502 
2526 
2550 

180 

2553 

2555 

2558 

2560 

2562 

2565 

2567 

2570 

2572 

2574 

181 
182 
183 

184 

185 
186 

187 
188 
189 

2577 
2601 
2625 

2648 
2672 
2695 

2718 
2742 
2765 

2579 
2603 
2627 

2651 
2674 
2697 

2721 
2744 
2767 

2582 
2605 
2629 

2653 
2676 
2700 

2723 
2746 
2769 

2584 
2608 
2632 

2655 
2679 
2702 

2725 

2749 

2772 

2586 
2610 
2634 

2658 
2681 
2704 

2728 
2751 
2774 

2589 
2613 
2636 

2660 
2683 
2707 

2730 
2753 
2776 

2591 
2615 
2639 

2662 
2686 
2709 

2732 
2755 

2778 

2594 
2617 
2641 

2665 
2688 
2711 

2735 

2758 
2781 

2596 
2620 
2643 

2667 
2690 
2714 

2737 
2760 
2783 

2598 
2622 
2646 

2669 
2693 
2716 

2739 
2762 
2785 

190 

2788 

2790 

2792 

2794 
2817 
2840 
2862 

2885 
2907 
2929 

2951 
2973 
2995 

2797 

2799 

2801 

2804 

2806 

2808 

191 
192 
193 

194 
195 
196 

197 
198 
199 

2810 
2833 
2856 

2878 
2900 
2923 

294.: 
2967 
2989 

2813 
2835 
2858 

2880 
2903 
2925 

2947 
2969 
2991 

2815 
2838 
2860 

2882 
2905 
2927 

2949 
2971 
2993 

2819 
2842 
2865 

2887 
2909 
2931 

2953 
2975 
2997 

2822 
2844 
2867 

2889 
2911 
2934 

2956 
2978 
2999 

2824 
2847 
2869 

2891 
2914 
2936 

2958 
2980 
3002 

2826 
2849 
2871 

2894 
2916 
2938 

2960 
2982 
3004 

2828 
2851 
2874 

2896 
2918 
2940 

2962 
2984 
3006 

2831 
2853 
2876 

2898 
2920 
2942 

2964 
2986 
3008 

200 

3010 

3012 

3015 

3017 

3019 

3021 

3023 

3025 

3028 

3030 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

250 


ADVANCED  ALGEBRA 


N. 

0 

1 

2     3 

4 

5 

6 

7 

8 

M 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

320r 

21 

22 
23 

24 
25 
26 

27 
28 
29 

3222 
3424 
3617 

3802 
3979 
4150 

4314 
4472 
4624 

3243 
3444 
3636 

3820 
3997 
4166 

4330 
4487 
4639 

3263 
3464 
3655 

3838 
4014 
4183 

4346 
4502 
4654 

3284 
3483 
3674 

3856 
4031 
4200 

4362 
4518 
4669 

3304 
3502 
3692 

3874 
4048 
4216 

4378 
4533 
4683 

3324 
3522 
3711 

3892 
40(35 
4232 

4393 
4548 
4698 

3345 
3541 
3729 

3909 
4082 
4249 

4409 
4564 
4713 

3365 
3560 
3747 

3927 
4099 
4265 

4425 
4579 

4728 

3385 
3579 
3766 

3945 
4116 
4281 

4440 
4594 
4742 

3404 
3598 
3784- 

3962 
4133 
4298 

4456 
4609 
4757 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 
32 
33 

34 
35 
36 

37 
38 
39 

4914 
5051 
5185 

5315 
5441 
5563 

5682 
5798 
5911 

4928 
5065 
5198 

5328 
5453 
5575 

5694 
5809 
5922 

4942 
5079 
5211 

5340 
5465 
5587 

5705 
5821 
5933 

4955 
5092 
5224 

5353 
5478 
5599 

6717 
5832 
5944 

4969 
5105 
5237 

5366 
5490 
5611 

5729 
5843 
5955 

4983 
5119 
5250 

5378 
5502 
5623 

5740 

6855 
5966 

4997 
5132 
6263 

5391 
5514 
5635 

5752 
5866 
5977 

6011 
6146 
5276 

5403 
5527 

6647 

6763 

5877 
6988 

5024 
5159 
5289 

5416 
6639 
6658 

5775 
5888 
5999 

5038 
5172 
5302 

5428 
5551 
5670 

6786 
6899 
6010 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 
42 
43 

44 
45 
46 

47 

48 
49 

6128 
6232 
6335 

6435 
6532 
6628 

6721 
6812 
6902 

6138 
6243 
6345 

6444 
6542 
6637 

6730 
6821 
6911 

6149 
6253 
6355 

6454 
6551 
6646 

6739 
6830 
6920 

6160 
6263 
6365 

6464 
6561 
6656 

6749 
6839 
6928 

6170 
6274 
6375 

6474 
6571 
6665 

6758 
6848 
6937 

6180 
6284 
6385 

6484 
6580 
6675 

6767 
6857 
6946 

6191 
6294 
6395 

6493 
6590 
6684 

6776 
6866 
6955 

6201 
6304 
6405 

6503 
6599 
6693 

6786 
6875 
6964 

6212 
6314 
6415 

6513 
6609 
6702 

6794 
6884 
6972 

6222 
6325 
6425 

6522 
6618 
6712 

6803 
6893 
6981 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

51 
52 
53 

54 
55 
56 

57 

58 
59 

7076 
7160 
7243 

7324 
7404 

7482 

7559 
7634 
7709 

7084 
7168 
7251 

7332 
7412 
7490 

7566 
7642 
7716 

7093 
7177 
7259 

7340 
7419 
7497 

7574 
7649 
7723 

7101 
7185 
7267 

7348 
7427 
7505 

7582 
7657 
7731 

7110 
7193 
7275 

7356 
7435 
7513 

7589 
7664 
7738 

7118 
7202 
7284 

7364 
7443 

7620 

7597 
7672 
7745 

7126 
7210 
7292 

7372 
7451 

7528 

7604 
7679 
7752 

7135 
7218 
7300 

7380 
7469 
7636 

7612 
7686 
7760 

7143 
7226 
7308 

7388 
7466 
7543 

7619 
7694 
7767 

7152 
7235 
7316 

7396 

7474 
7551 

7627 
7701 
7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7826 

7832 

7839 

7846 

61 
62 
63 

64 
65 
66 

67 
68 
69 

7853 
7924 
7993 

8062 
8129 
8195 

8261 
8325 
8388 

7860 
7931 
8000 

8069 
8136 
8202 

8267 
8331 
8395 

7868 
7938 
8007 

8075 
8142 
8209 

8274 
8338 
8401 

7875 
7945 
8014 

8082 
8149 
8215 

8280 
8344 
8407 

7882 
7952 
8021 

8089 
8156 

8222 

8287 
8351 
8414 

7889 
7959 
8028 

8096 
8162 
8228 

8293 
8367 
8420 

7896 
7966 
8036 

8102 
8169 
8235 

8299 
8363 
8426 

7903 
7973 
8041 

8109 
8176 
8241 

8306 
8370 
8432 

7910 
7980 
8048 

8116 
8182 
8248 

8312 
8376 
8439 

7917 
7987 
8055 

8122 

8189 
8254 

8319 
8382 
8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

LOGARITHMS 


251 


N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 
72 
73 

8513 
8573 
8633 

8519 
8579 
8639 

8525 

8585 
8645 

8531 
8591 
8651 

8537 
8597 

8657 

8543 
8603 
8663 

8549 
8609 
8669 

8555 
8615 
8675 

8561 
8621 
8681 

8667 
8627 
8686 

74 

75 
76 

8692 
8751 
8808 

8698 
8756 
8814 

8704 
8762 
8820 

8710 
8768 
8825 

8716 

8774 
8831 

8722 
8779 
8837 

8727 
8785 
8842 

8733 
8791 
8848 

8739 
8797 
8854 

8745 
8802 
8859 

77 
78 
79 

8865 
8921 
8976 

8871 
8927 
8982 

8876 
8932 
8987 

8882 
8938 
8993 

8887 
8943 
8998 

8893 
8949 
9004 

8899 
8954 
9009 

8904 
8960 
9015 

8910 
8965 
9020 

8915 
8971 
9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 
82 
83 

9085 
9138 
9191 

9090 
9143 
9196 

9096 
9149 
9201 

9101 
9154 
9206 

9106 
9159 
9212 

9112 
9165 
9217 

9117 
9170 
9222 

9122 
9175 
9227 

9128 
9180 
9232 

9133 
9186 
9238 

84 
85 
86 

9243 
9294 
9345 

9248 
9299 
9350 

9253 
9304 
9355 

9258 
9309 
9360 

9263 
9315 
9365 

9269 
9320 
9370 

9274 
9325 
9375 

9279 
9330 
9380 

9284 
9335 
9385 

9289 
9340 
9390 

87 
88 
89 

9395 
9445 
9494 

9400 
9450 
9499 

9405 
9455 
9504 

9410 
9460 
9509 

9415 
9465 
9513 

9420 
9469 
9518 

9425 
9474 
9523 

9430 
9479 
9528 

9435 
9484 
9533 

9440 
9489 
9538 

90 

9542 

9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

91 
92 
93 

9590 
9638 
9685 

9595 
9643 
9689 

9600 
9647 
9694 

9605 
9652 
9699 

9609 
9657 
9703 

9614 
9661 
9708 

9619 
9666 
9713 

9624 
9671 
9717 

9628 
9675 
9722 

9633 
9680 
9727 

94 
95 
96 

9731 
9777 
9823 

9736 

9782 
9827 

9741 
9786 
9832 

9745 
9791 
9836 

9750 
9795 
9841 

9754 
9800 
9845 

9759 
9805 
9850 

9763 
9809 
9854 

9768 
9814 
9859 

9773 
9818 
9863 

97 

98 
99 

9868 
9912 
9956 

9872 
9917 
9961 

9877 
9921 
9965 

9881 
9926 
9969 

9886 
9930 
9974 

9890 
9934 
9978 

9894 
9939 
9983 

9899 
9943 
9987 

9903 
9948 
9991 

9908 
9952 
9996 

100 

0000 

0004 

0009 

0013 

0017 

0022 

0026 

0030 

0035 

0039 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

216.  Exponential  equations.    Equations  in  which  the  variable 
occurs  only  in  the  exponents  may  often  be  solved  by  the  use  of 
tables  of  logarithms  if  one  keeps  in  mind  the  fact  that 
log  oc^  =  ic  log  a. 


EXERCISES 

Solve  the  following : 

1.  10^-1  =  4. 

Solution :    Taking  the  logarithm  of  both  sides  of  the  equation,  we  have 
(x-1)  log  10  =  log  4, 
or  since  log  10  =  1,    a;  =  log  4  +  1  =  .G021  +  1  =  1.0021. 


252  ADVANCED  ALGEBRA 

2.  4* -3^  =  8, 
2* .  81/  =  9. 

Solution :   Taking  the  logarithms  of  the  equations,  we  have 

a;log4  +  ylogS  =  log  8, 
a:  log  2  +  y  log  8  =  log  9, 

or  jc21og2  4-ylog3  =  31og2,  (1) 

a;log2 +  y31og2  =  21og3. 
Eliminate  x. 

x21og2+     2/log3  =  31og2 

a;21og2  +  y61og2  =  41og3 

2/(log3  -  61og2)  =  31og2  -  41og3 

_31og2-41og3_3  X  .3010-4  x  .4771 
^~   Iog3-01og2    ~      .4771  -  6  X  .3010 

.9030  -  1.9084  _  -  1.0054  _  1.0054 
~  .4771  -  1.8060  ~  -  1.3289  ~  1.3289* 

Perform  this  division  by  logarithms. 

log  1.0054  =  10.0023  -  10 
log  1.3289=      .1235 

logy=    9.8788-10 
y=      .7565. 
Substituting  in  (1), 

a;  =  31og2  -  .7565 log3  _  .9030  -  .7565  x  .4771 
2  log  2  ~  .6020 

Compute  .7666  x  .4771  by  logarithms. 

log.  7565=    9.8788-10 
log. 4771=    9.6786-10. 
log  result  =  19. 5574  -  20 
result  =     .3609. 

„  .9030 -.3609      .5421 

Hence  x  — = 

.6020  .6020 

log.5421  =  19.7341  -20 

log. 6020=    9.7796  -  10 

log;c=    9.9545-  10 

X  =     .9005. 

3.  6*  =  2.  4.  4y  =  3.  5.  7*+«  =  5. 

6.  32x  +  i_5.  7.4^-1  =  6^+1.  8.  22a:+3_ 6^-1  =  0. 

a^.6y  =  m,  -^    a^.6!'  =  w, 

c*  .  d?/  =  71.  '  x-\-  y  —  n. 


LOGARITHMS  253 

--    2^.  2^  =  222,  -2    a2^-3.a3y-2:=a8. 

'  x-y  =  A.  8x  +  2y  =  n. 

-  3    3*  •  4y  =  15562,  ..     ^^2^^  .  ^/^^F^i  =  a\ 

4^  .  52/ =  128000.  4,^:,— ^     3 


^l)3x  +  5  -  ^sjb^y  + 1  =  610. 

217.  Compound  interest.  If  |1500  is  at  the  yearly  interest 
of  3%,  the  total  interest  for  a  year  is  $1500  •  (0.03)  =  $45.  The 
total  sum  invested  at  the  end  of  a  year  would  be  $1545. 

Let,  in  general,  P  represent  a  sum  of  money  in  dollars. 

Let  r  represent  a  yearly  rate  of  interest. 

Then  P  •  r  represents  the  yearly  interest  on  P,  and 
P-f  p.r  =  P(r  +  1) 
represents  the  total  investment,  principal  and  interest,  at  the 
end  of  a  year. 

Similarly,  P  (r  +  1)  r  is  the  second  year's  interest,  and 

P{r  +  l)r  +  P(r  +  1)  =  P(r2  +  r  +  r  +  1)  =  ^(^  +  1)^ 
is  the  total  investment  at  the  end  of  two  years. 

In  general,  A  =  P  (r  +  1)"  (1) 

is  the  total  accumulation  at  the  end  of  tz  years.  If  we  know  r,  P, 
and  n,  we  can  by  (1)  find  A.  If  we  take  the  logarithm  of  both 
sides  of  the  equation,  we  have 

log  A  =  log  P  -\-  n  log  (r  +  1), 

log^-logP^  (2) 

log(r  +  l)  ^  ^ 

Hence  if  we  know  A,  P,  and  r,  we  can  find  n. 
If  the  interest  is  computed  semiannually,  we  have  as  interest 

v 

at  the  end  of  a  half-year  P  •  ^ '   while  the  entire  sum  would  be 

P  (-  4- 1 ) .  Reasoning  as  above,  we  find  that  if  the  interest  is  com- 
puted semiannually,  the  accumulation  at  the  end  of  n  years  is 

^=pg  +  l)"  (3) 

Similarly,  n=  l"g -^  '  l»g  f .  (4) 


logg  +  l) 


254  ADVANCED  ALGEBRA 


3 


If  the  interest  is  computed  k  times  a  year,  we  liave  at  the  end 
of  n  years 

A-P(^  +  1)    ,  (5) 


Hogg  +  l) 


EXERCISES 

In  such  exercises  as  the  following,  four-place  tables  are  not  sufficiently 
exact  to  obtain  perfect  accuracy.  In  general,  the  longer  the  term  of  years 
and  the  more  frequent  the  compounding  of  interest,  the  greater  the  inaccuracy. 

1.  If  $1600  is  placed  at  3|%  interest  computed  semiannually  for  13  years, 
to  how  much  will  it  amount  in  that  time  ? 


Hence 


ion:   By  formula  (3), 

A  = 

4 

-) 

2n 

P  =  1600,  r  = 

.03|, 

n  = 

13 

36           A  =  1600  {-^ 

\26 

1600 

\400 

-)^=-Q^- 

log  1600=    3.2041 

26  log407  =  67.8496 

71.0537 

261og400  =  67.6546 

log^=    3.3991 

A  =  $2507. 

Iog407  =  2.e096 

26 

156576 

52192 

67.8496 

log  400  =2.6021 

26 

156126 

52042 

67.6546 

2.  After  how  long  will  $600  at  6%  computed  annually  amount  to  $1000  ? 

Solution :   By  formula  (2)  we  have 

_  log  J.  —  logP 

log(r  +  l) 

A  =  1000,  P  =  600,  r  =  .06. 

log  1000  -  log  600      3  -  2.7782       .2218      _  ^,. 

n  =  -^ = =^ =  8./0  years. 

log  1.06  .0253  .0253  ^ 

.76  year  =  .76  •  12  =  9.12  months. 

.12  month  =  .12  •  30  =  3.6  days. 

Thus  n  =  8  years  9  months  3.6  days. 


LOGARITHMS  255 

In  the  following  exercises  the  interest  is  computed  annually  unless  the 
contrary  is  stated. 

3.  To  what  will  $3750  amount  in  20  years  if  left  at  6%  interest  ? 

4.  To  what  sum  will  $25,300  amount  in  10  years  if  left  at  5%  interest 
computed  semiannually? 

5.  To  what  does  $1000  amount  in  10  years  if  left  at  6%  interest  computed 
(1)  annually,  (2)  semiannually,  (3)  quarterly  ? 

6.  A  sum  of  money  is  left  22  years  at  4%  and  amounts  to  $17,000.    How 
much  was  originally  put  at  interest  ? 

7.  What  sum  of  money  left  at  4|%  for  30  years  amounts  to  $30,000  ? 

8.  What  sum  of  money  left  10  years  at  4i%  amounts  to  the  same  sum  as 
$8549  left  7  years  at  5%  ? 

9.  If  a  man  left  a  certain  sum  11  years  at  4%,  it  would  amount  to  $97  less 
than  if  he  had  left  the  same  sum  9  years  at  5%.    What  was  the  sum  ? 

10.  Which  yields  more,  a  sum  left  10  years  at  4%  or  4  years  at  10%? 
What  is  the  difference  for  $1000  ? 

11.  Two  sums  of  money,  $25,795  in  all,  are  left  20  years  at  4f%.  The 
difference  in  the  sums  to  which  they  amount  is  $14,660.    What  were  the  sums  ? 

12.  At  what  per  cent  interest  must  $15,000  be  left  in  order  to  amount  to 
$60,000  in  32  years  ? 

13.  At  what  per  cent  must  $3333  be  left  so  that  in  24  years  it  will 
amount  to  $10,000  ? 

14.  Two  sums  of  which  the  second  is  double  the  first  but  is  left  at  2%  less 
interest  amount  in  36 1  years  to  equal  sums.  At  what  per  cent  interest  was 
each  left  ? 

15.  In  how  many  years  will  a  sum  double  if  left  at  5%  interest  ? 

16.  In  how  many  years  will  a  sum  double  if  left  at  6%  interest  computed 
semiannually  ? 

17.  In  how  many  years  will  a  sum  amount  to  ten  times  itself  if  left  at 
4%  interest  ? 

18.  In  how  many  years  will  $17,000  left  at  4^%  interest  amount  to  the 
same  as  $7000  left  at  5|%  for  20  years  ? 

19.  On  July  1,  1850,  the  sum  of  $1000  was  left  at  4i%  interest.  When 
paid  back  it  amounted  to  $2222.    When  did  this  occur  ? 

20.  Prove  formulas  (1),  (3),  and  (5)  by  complete  induction. 


CHAPTEE  XXI 
CONTINUED  FRACTIONS 

218.  Definitions.   A  fraction  in  the  form 

e-¥f 

where  a,  &,•••,  ^,  ••  •  are  real  numbers,  is  called  a  continued  fraction. 
We  shall  consider  only  those  continued  fractions  in  which  tlie 
numerators  b,  d,  /,  etc.,  are  equal  to  unity  and  in  which  the 
letters  represent  integers,  as  for  example  : 

»!  H .  written  a^-\ —      — 

When  the  number  of  quotients  ^2?  ^s?  ^4?  •  •  •  is  finite  the  frac- 
tion is  said  to  be  terminating.  When  the  fraction  is  not  terminat- 
ing it  is  infinite.  We  shall  see  that  the  character  of  the  numbers 
represented  by  terminating  fractions  differs  widely  from  that  of 
the  numbers  represented  by  infinite  continued  fractions.  We  shall 
find,  in  fact,  that  any  root  of  a  linear  equation  in  one  variable, 
i.e.  any  rational  number,  may  be  represented  by  a  terminating 
continued  fraction,  and  conversely;  furthermore,  that  any  real 
irrational  root  of  a  quadratic  equation  may  be  represented  by  the 
simplest  type  of  infinite  continued  fractions,  and  conversely. 

219.  Terminating  continued  fractions.  If  we  have  a  terminat- 
ing continued  fraction,  where  ^i,  a2j  •  •  •  are  integers,  it  is  evident 
that  by  reducing  to  its  simplest  form  we  obtain  a  rational  num- 
ber.   The  converse  is  also  true,  as  we  can  prove  in  the  following 

256 


CONTINUED  FRACTIONS  257 

Theorem.    Any  rational  number  may  he  expressed  as  a  ter- 
minating fraction. 

Let  -  represent  a  rational  number.    Divide  a  by  J,  and  let  a^ 

be  the  quotient  and  c  (which  must  be  less  than  h)  the  remainder. 
Then  (§  26)  , 

-^a,  +  -  =  a,  +  j 


Divide  h  by  c,  letting  ^g  t)®  the  quotient  and  d  (which  must  be 
less  than  c)  the  remainder.    Then 

-  =  «!  H 

b  a^  -\-  d 

c 

Continuing  this  process,  the  maximum  limit  of  the  remainders 
in  the  successive  divisions  becomes  smaller  as  we  go  on,  until 
finally  the  remainder  is  zero.  Hence  the  fraction  is  terminating. 
It  is  noted  that  the  successive  quotients  are  the  denominators  in 
the  continued  fraction. 

EXERCISES 

1.  Convert  the  following  into  continued  fractions 

(a)  ^VV- 

Solution:   247|77[0 

77J247|3 
231 
16J77[4 
64 

13J16[1 
13 

3J13[4 
12 
1J313 
3 
0 
The  continued  fraction  is 

ZL-1      1      1      1     1. 

247  ~  3  +  4  +  1  +  4  +  3* 


258  ADVANCED  ALGEBRA 

(b)  if.  (c)  ^Vt.  (d)  m- 

(e)  iM-  (f)  Iff.  (g)  /jVf- 

(h)  HH-  (i)  Hf.  (J)  mh 

2.  Express  the  following  continued  fractions  as  rational  fractious. 

^''^2  +  3  ^"^l  +  r  ^"^-a  +  l- 

(d)l  I.  (e)i  11.         (f)l  1  1. 

^  '  a;  +  x  ^  M+2  +  3         ^  '   3  +  4  +  5 

,.11111  ,^111111 

(g)  ----- .  (h)  ------  . 

^^2+4+2+4+2  ^^1+2+3+1+2+3 

220.  Convergents.  The  value  obtained  by  taking  only  the  first 
n  ~  1  quotients  in  a  continued  fraction  is  called  the  nth.  convergent 
of  the  fraction. 

Thus  iu  the  fraction 

1+1111 

2+3+2+6  : 

1  is  the  first  convergent, 

1       3 

1  +  -  =  -  is  the  second  convergent, 

■a        2t 

l  +  -=l+-  =  —  is  the  third  convergent,  etc. 

3 
When  there  is  no  whole  number  preceding  the  fractional  part  of  the  continued 
fraction  the  first  convergent  is  zero.    Thus  in 

111 

2  +  3  +  5 
\  is  called  the  second  convergent. 

In  the  continued  fraction 

,1111 

^2  +  ^3  +  ^4  +  «5  H 

let  -i,  -1,  -i,  . ..  represent  the  successive  convergents  expressed 

9.\    ?2    ^z 
as  rational  fractions. 

Then  for  the  first  convergent  we  have 

^  =  — >  or^i  =  ai,  g'i  =  l. 
9.\ 


CONTINUED  FRACTIONS  259 

For  the  second  convergent  we  have 

,    1       a^a^  + 1      J92  ,  H  ,  ^ 

«!  H =  =  — >    or  ^2  =  «^2<^l  +  1  =  «2i?i  +  1, 

^2  <^2  2'2 

!Z2  =  «2  =  ^22'l- 

For  the  third  convergent  we  have 

^2  +    1^  a3«^2  +  1  «^3«2  +1  2'3 

«3 

or  i?3  =  %  (^2%  +  1)  4-  ^1  =  ^3^2  +i?l, 

?Z3  =  «3«2  +  1  =  %'Z2  +  (1\- 

This  indicates  that  the  form  of  the  r^th  convergent  is 

9'n  a«2'n-l  +  2'»-2  ^ 

This  is  in  fact  the  case,  as  we  proceed  to  show  by  complete 
induction. 

We  have  already  established  form  (1)  for  n  =  2  and  n  =  ^. 
We  assume  it  for  n  =  m,  and  will  show  that  its  validity  for 
n  =  m  -\-l  follows.    The  (m  +  l)th  convergent  differs  from  the 

mth  only  in  the  fact  that  a^  -\ appears  in  the  continued 

fraction  in  place  of  a^.    In   (1)    replace   n   by  m,  and    a„  by 

a^  H J     and  we  have 


Pr 


1+1  __  \ *^m+l/ 

'■^^  Um  + )  2'm-l  +  2'm-2 

\  «'m+l/ 

_  (Q^m-H<^>n  +  ^)Pm^l  +  Q^„.  +  1  J^m  -  2 
(am+;iam  +  l).?m-l  +  «m  +  l?m-2 

_Q^m4-lKi>m-l+i>m~2)+i>»»-l 
«^».  +  l(am2'm-l  +  9'm-2)+  2'm-l 


which  is  form  (1). 


260  ADVANCED  ALGEBRA 

EXERCISES 

1.  Express  the  following  as  continued  fractions,  and  find  the  convergents. 

(a)  !?. 

Solution :   By  the  method  already  explained,  we  find  that 

30  _1      1      1      1      1      1 

il~l  +  2+l  +  2  +  l  +  2' 
Here  ai  =  0,  a2  =  1,  as  =  2,  04  =  1,  as  =  2,  aQ  =  1,  a?  =  2. 
The  first  convergent  is  evidently  0,  the  second  is  1,  and  the  third  is 
1  2 


mi-      ir  ^.  ^   -      P4         CliPS  +P2         1-2+1         3 

The  fourth  convergent  is  —  =  -^^ — —  = =  -. 

Qi      a^qs  +  ga       1-3  +  1      4 

mu    ^x^u  .  •    P5      ttsP*  +  P3      2-3  +  2       8 

The  fifth  convergent  is  —  =  —^ — ^  = =  —  • 

?5      a^qi  +  ^3       2-4  +  3      11 

aePn  +  P4        1-8  +  3        11 


The  sixth  convergent  is  ^^  = 
The  seventh  convergent  is 


^6      tteO's  +  5'4       1-11  +  4      15 
Pi      OtPg+Ps       2-11  +  8       30 


(b)  3¥2.  (c)  If.  (d)  t¥t-  (e)  tVt-  (f)  t?7- 


Qi      a^q&  +  q^      2-15  +  11      41 

(f)  \'-'- 

(g)TVj.  (b)TW^.  (i)?\V  U)  mV 

2.  Find  the  value  of  the  following  by  finding  the  successive  convergents. 

,,11111  ,^,11111 

(a)  -   -   -   -   -•  (b)  -   -   -   -   -. 

^^2+1+2+1+2  ^'3+2+3+2+3 

(c^  1   i   i   i   i   i  (d^  1   1   1   1   1   1 

^^^2  +  3+1  +  1  +  3  +  2*  ^^3  +  3_^34.34.34.3' 

(e^  1  1 '  1  1   1  1  (f\  1  1   1  1  1  1 

^^^6  +  3  +  1  +  1  +  3  +  6'  ^^1  +  34.5+5  +  3+1' 

1     1     1  1   1   1 

^^^  (X  -  1)  +  X  +  (X  +  1)  *  ^  ^  X  +  X  +  x' 

221.  Recurring  continued  fractions.  We  have  seen  that  e very- 
terminating  continued  fraction  represents  a  rational  number,  and 
conversely.  We  now  discuss  the  character  of  the  numbers  repre- 
sented by  the  simplest  infinite  continued  fractions.  A  recurring 
continued  fraction  is  one  in  which  from  a  certain  point  on,  a 
group  of  denominators  is  repeated  in  the  same  order. 


CONTINUED  FRACTIONS  261 

-----  ^ 

^^  3  +  2  +  3  +  2  +  3+2  +  *"' 

111111 

1+2+3+1+2+3+'" 

are  recurring  continued  fractions  if  the  denominators  are  assumed  to  repeat 
indefinitely  as  indicated. 

That  a  repeating  continued  fraction  actually  represents  a  num- 
ber we  shall  establish  in  §  223.  Unless  this  fact  is  proven,  one 
runs  the  risk  of  dealing  with  symbols  which  have  no  meaning. 
If  for  certain  continued  fractions  the  successive  convergents 
increase  without  limit,  or  take  on  erratic  values  that  approach  no 
limit,  it  is  important  to  discover  the  fact.  All  the  fractions  that 
we  discuss  actually  represent  numbers,  as  we  shall  see. 

We  shall  consider  only  continued  fractions  in  which  every 
denominator  has  a  positive  sign. 

Theorem.  Every  recurring  continued  fraction  is  the  root  of  a 
quadratic  equation. 

T   w     •     .  111111 

Let,  tor  instance,         a;  =  -      -      -      -      -      -      •••. 

'  '  a -\-  b -\-  c -\- a -\-  b -\-  G  -[- 

Evidently  the  part  of  the  fraction  after  the  first  denominator  c 
may  be  represented  by  cc,  and  we  have  thus  virtually  the  termi- 
nating fraction 


X 


a  -\-  b  -\-  c  -\-  X 


The  second  convergent  is  — 
The  third  convergent  is 

ab  -\-l      q^ 
The  fourth  convergent,  or  a?,  gives  us 

^jP4^CT4^3+-^2^        (c+-a;)5+-l 
q^      ci^qz  4-  S'2       (c  +  x)  (ab  +- 1)  4-  fit 
Simplifying,  we  get 

(ab  +  l)x^-{- lc(ab  -{-l)-{- a  -  b'jx  -  be -1  =  0, 


262  ADVANCED  ALGEBRA 

which  is  a  quadratic  equation  whose  root  is  a?,  the  value  of  the 
continued  fraction. 

Since  this  equation  has  a  negative  number  for  its  constant  term 
it  has  one  positive  and  one  negative  root.  The  continued  fraction 
must  represent  the  positive  root,  since  we  assume  that  the  letters 
la,  h,  c  represent  positive  integers.  The  quadratic  equation  whose 
root  is  a  recurring  continued  fraction  with  positive  denomina- 
tors will  always  have  one  positive  and  one  negative  root.  The 
equation  will  be  quadratic,  however,  whatever  the  signs  of  the 
denominators  may  be. 

The  proof  may  be  extended  to  the  case  where  there  are  any 
number  of  recurring  denominators  or  any  number  of  denominators 
before  the  recurrence  sets  in.  Since  every  real  irrational  root 
of  a  quadratic  equation  is  a  surd,  our  result  is  equivalent  to  the 
statement  that  every  recurring  continued  fraction  may  be  ex jft*essed 
as  a  surd. 

EXERCISES 

Of  what  quadratic  equations  are  the  following  roots  ?  Express  the  con- 
tinued fraction  as  a  surd. 

ill      1      1 

*  2  +  3  +  2  +  3  +  ""' 

Solution :   Let  x  =  -      -         . 

2  +  3  +  a: 

Then  x  =  l      1  ^  +  ^ 


2+-  6+2X+1 

3  +  « 


or  2  0:2  +  6  X  -  3  =  0. 

Solving  this  equation,  we  get 

-  3  +  VT6                 -  3  -  Vl5 
X\  = or  X2  = 

2  2 

Since  x^  is  negative,  xi  must  be  the  surd  that  is  represented  by  the  con- 
tinued fraction. 

1111  _3  +  Vl6 


Thus 
2. 


2+3+2+3+  2 

1111  3IIII 

1  +  2  +  1  +  2  +  "*'  ■  3  +  2  +  3  +  2  + 


6.2  +  1      1      1      1      . 

2+1+2+1+ 

••• 

m.r.u.  =  2  +  l^\^. 

••. 

then         ^-^=l^\^-- 

and          .  =  2  +  i^-;^(^. 

-2) 

CONTINUED  FRACTIONS  263 

^111111  silil 

•   1  +  2  +  3  +  1+2  +  3  +  ""  •3+1  +  3  +  1  +  *"' 

«      o  1  1  1 

7.  3  +  -     -     -      .... 

^3+4+5+ 

8.1  +  1      1      1      1 

3+4+3+4+ 

9.  1  +  i     1     1     i      i      1      .... 

1+2+3+1+2+3+ 

10.  i     1     1    -1     1     i     .... 

2+1+2+2+1+2+ 

222.  Expression  of  a  surd  as  a  recurring  continued  fraction. 

This  is  the  converse  of  the  problem  discussed  in  the  last  sec- 
tion, and  shows  that  recurring  continued  fractions  and  quadratic 
equations  are  related  in  the  same  intimate  way  that  terminat- 
ing fractions  and  rational  numbers  (i.e.  the  roots  of  linear  equa- 
tions) are  connected.  We  seek  to  express  an  irrational  number, 
as,  for  instance,  V2,  as  a  continued  fraction.  This  we  may  do 
as  follows. 

Since  1  is  the  largest  integer  in  V2  we  may  write 

V2  =  l  +  (V2-l)  =  l+(-^^^^. 
Rationalizing  the  numerator,  we  have 

V2  =  i  +  --i — 

V2+-1 

Since  2  is  the  largest  integer  in  V2  +- 1  we  have 

V2  =  l+ 7^= 7  =  1  + 


2+-(V2-l)  o  I  (V2-I) 

Rationalizing  the  numerator  V2  —  1,  we  have 

^"^viTI       ^  +  2+(V2-i) 


264  ADVANCED  ALGEBRA 

By  continuing  this  process  we  continually  get  tlie  denomi- 
nator 2.    Thus  11 

This  process  consists  of  the  successive  application  of  two  opera- 
tions, and  affords  the 

EuLE.  Express  the  surd  as  the  sum  of  two  numbers  the  first 
of  which  is  the  largest  integer  that  it  contains. 

nationalize  the  numerator  of  the  fraction  whose  numerator  is 
the  second  of  these  numbers.  Repeat  these  operations  until  a 
recurrence  of  denominators  is  observed. 

This  process  may  be  applied  to  any  surd,  and  a  continued  frac- 
tion which  is  recurring  will  always  be  obtained.  We  shall  con- 
tent ourselves  with  a  statement  of  this  fact  without  proof. 

If  the  surd  is  of  the  form  a  —  V^,  a  continued  fraction  may  be 
derived  for  +  V^  and  its  sign  changed.  Since  the  real  roots  of 
any  quadratic  equation  cc^  +  2  a^x  +  ag  =  0  are  surds  of  the  form 
a  ±  V^,  where  a  and  b  are  integers,  it  appears  that  the  roots  of 
any  such  equation  may  be  expressed  as  recurring  continued  frac- 
tions. It  can  be  shown  that  the  real  roots  of  the  general  quad- 
ratic equation  a-o^^  +  ajcc  -f  ctg  =  0  may  also  be  so  expressed. 

EXERCISES 

1.  Express  the  following  surds  as  recurrent  continued  fractions, 
(a)  2  +  Vs. 
Solution : 

2+V3  =  3  +  (V3-l)  =  3+  (1) 

=  34^^  =  3+       2' 


V3+    1  V3+I 

=  3  +  ^^^ —  =  3  + J =  3  + 


V3  +  I  ^      V3+I  _ .  V3-I 

2  .2  "^2 

3  +  1^       3-1      =3  +  ^^^  =  3  +  1      1 

2(V3  +  1)  V3  +  I  2-|-(V3-l) 

1 


conti:nued  fkactions  265 

But  since  Vs  —  1  is  the  same  number  that  we  have  in  (1),  this  fraction 
repeats  from  this  point  on,  and  we  have 

2+V3  =  8  +  l      1      I      1      .... 
1+2+1+2+ 


(b)  V5. 

(c)  Vl7. 

(d)  V65. 

(e)  V47. 

(f)  Vli. 

(g)  V23. 

(h)   V3i. 

(i)  Vl9. 

(J)  V62. 

(k)   V79. 

(1)   V98. 

(m)  V88. 

(n)  V22. 

(0)  Vis. 

(p)  V59. 

(q)  VlOl. 

(r)  7  +  VII. 

(s)  8- 

-V3 

(t) 

3-V23. 

2.  Express  as  a  continued  fraction  the  roots  of  the  following  equations, 
y   (a)  a;2  -  7  X  -  3  =  0.  /  (b)  x^  +  2  a;  -  6  =  0. 

(c)  x2  +  3x  -  8  =  0.  (d)  x2  -  4x  -  4  =  0. 

223.  Properties  of  conver gents.  The  law  of  formation  of  con- 
vergents  given  in  §  220  is  valid  whether  the  continued  fraction 
is  terminating  or  infinite.  We  should  expect  that  in  the  case  of 
an  infinite  fraction  the  successive  convergents  would  give  us  an 
increasingly  close  approximation  to  the  value  of  the  fraction. 
This  is  indeed  the  fact,  as  we  shall  see. 

Theorem.  The  difference  between  the  nth  and  (n  +  l)st  con- 
vergents is  ^ 

Mn  +  l 

We  prove  this  theorem  by  complete  induction. 
Let  the  continued  fraction  be 


a,  H —      — 

«2  +  ^3  +  »4  + 

Then  the  first  and  second  convergents  are  respectively 
Then  £2-a=(^i^L±l)_„,  =  i. 

2'2  S'l  »2  «2 


(1) 


266  ADVANCED  ALGEBRA 

Since  ^i  =  1?  2'2  =  ^2? 

we  have  ^zl±i  _^  =  izL^yHl  for  n  =  1. 

We  assume  that  the  theorem  holds  for  n  —  m^  that  is, 

Prn±l  _  ^    -y^m^m  +  i+gmPm  +  l  ^  {zlJ}!^. 

We  must  prove  that  it  holds  for  n  =  m  -{•  1, 

Now  since  -^"^^^  —  -^'"  +  ^  ^Pm  +  l^m  +  2  YPm  +  29^m+l^ 
2'm  +  2  S'm  +  l  2'm  +  l2'»i  +  2 

our  theorem  reduces  to  proving  that  the  numerator 

--  Pm+lQm  +  2  ■^Pm  +  2^m  +  l=(-  l)""^"-  (2) 

In  the  left-hand  member  of  (2)  set 

«m  +  2'7m+l  +  ^m   =  2'm  +  2,  (1)>    §  220 

and  «^m  +  2i^m  +  l+i^m=I>m  +  2- 

Then       -- Pm+li^m+^^m  +  l  H-  2'm)f  (^m+2P„.+  l  +i?».)?n.  +  l 
=^m+l^m-^Pm<Im+l  =  -  (Pm'Ln+ I  -  Pm  +  l^m) 

Corollary  I.  The  difference  between  the  successive  convergents 
of  a  continued  fraction  with  positive  denominators  approaches 
zero  as  a  limit. 

Since  q^  =  a^q„_-^  +  q^-^,  evidently  q^  increases  without  limit 
when  n  is  increased,  since  to  obtain  a^  we  add  together  positive 
numbers  neither  one  of  which  can  vanish. 

Thus  we  can  find  a  value  of  n  large  enough  so  that   —  j  and 

1  .  ^«     . 

hence  ?   will  be  smaller  than  any  assigned  number,  which 

is  another  way  of  stating  that  as  n  increases approaches 

zero  as  a  limit.  ^'•^'*  "*"  ^ 


CONTINUED  FliACTIONS  267 

Corollary  II.    The  even  convergents  decrease,  while  the  odd 
convergents  increase,  as  n  increases. 

We  must  show  that 

PTn  +  2  Pm 

9.m  +  2  5'm 

is  negative  or  positive  according  as  ni  is  even  or  odd.    Adding 
and  subtracting  ?  we  have 


?/m  +  1 

Pm  +  2  Pm  (Pjn±2  Pin±\  i     , 

m+l 


2'm  +  2  ^m  Vlm  +  2  <! 


/Pjn±l_ln\ 
X^m  +  l  ^mj 


_  (-1)"»+^  (-1)"*  +  ^ 

By  Corollary  I,  the  denominator  of  the  first  fraction  exceeds 
that  of  the  second.  Hence  when  m  is  odd  the  sum  in  the  last 
member  of  the  equation  is  positive,  and  when  7n  is  even  the  sum 
is  negative. 

We  now  se*  that  any  recurring  fraction  of  the  type  considered 
in  §  221  actually  represents  a  number  in  the  sense  of  §  74.  We 
have  seen  that  the  successive  odd  convergents  continually  increase, 
while  the  even  convergents  continually  decrease,  until  the  differ- 
ence between  a  pair  of  them  is  very  small.  Such  sequences  of 
numbers  we  have  seen  (§73  ff.)  define  real  numbers. 

224.  Limit  of  error.  We  are  now  in  a  position  to  state  a  maxi- 
mum value  for  the  error  made  in  taking  any  convergent  of  a  con- 
tinued fraction  for  the  fraction  itself. 

Theorem.  The  maximum  limit  of  error  in  taking  the  nth 
convergent  for  the  continued  fraction  is  less  than 

Since  by  the  theorem  of  the  last  section  the  value  of  the  frac- 
tion is  between  any  pair  of  consecutive  convergents,  it  must 
differ  from  either  of  these  convergents  by  less  than  they  differ 

from  each  other,  that  is,  by  less  than 


268  ADVANCED  ALGEBRA 

EXERCISES 
Find  a  convergent  that  differs  by  less  than  .001  from  each  of  the  following ; 

1.    V6. 
Solution : 

V6  =  2  +  ( V6  -  2)  =  2  +    ^_~^    =  2  + 


V6  +  2  V6'+2 

2 

=  2  +  i  6-4       =2  +  5  1       =^  +  lj_l 

Since  the  last  surd  repeats  the  one  in  the  first  equation  we  have 

V6  =  2  +  i      1      1      1      .... 

2+4+2+4+ 


Since 


Pi  _  2  .     J92  _  5 .     Pa  _  4  •  5  +  2  _  22  . 
gi  ~  1 '     52  ~  2  '     ^3      4  .  2  +  1  ~  9  ' 

P4      2-22  +  5      49.     ps      4-49  +  22 
g4~2.9  +  2~20'     95~4-20  +  9~ 

218 
89 

1-^-1    <.001. 

54^5      20  -  89      1780 

we  see  by  §  224  that  |f  satisfies  the  condition,  of  the  problem. 

2.   V7.  3.   V46.  ,  4.   Vs.  5.   Vl9. 

6.  V36.  7.  V32.  8.   V6l.  9.  V65. 

10.  S+V2.-        11.   V99.  12.   Vil.  13.   Vl3. 

14.  The  number  ir  has  the  value  3. 14169.  Find  by  the  method  of  continued 
fractions  a  series  of  convergents  the  last  of  which  differs  from  this  value  by 
less  than  .OCioi. 


CHAPTEE  XXII 

INEQUALITIES 

225.  General  theorems.  We  say  that  a  is  greater  than  h  when 
a  —  6  is  positive.  If  a  —  h  is  negative,  then  a  is  less  than  h. 
Thus  any  positive  number  or  zero  is  greater  than  any  negative 
number.  As  we  distinguished  between  identities  and  equations 
of  condition  in  §  53,  so  in  this  discussion  we  observe  that  some 
statements  of  inequality  are  true  for  any  real  value  of  the  letters, 
while  others  hold  for  particular  values  only.  The  former  class 
may  be  called  unconditional  inequalities,  the  latter  conditional. 

Thus  a2  >  —  1  is  true  for  any  real  value  of  a  and  is  unconditional,  while 
a;  —  1  >  2  only  when  a;  is  greater  than  3  and  is  consequently  conditional. 

The  two  inequalities  <*  >  5,  g'>  d  are  said  to  have  the  same 
sense.  Similarly,  a  <b,  c  <d  have  the  same  sense.  The  inequal- 
ities a>h,  c  <id  have  a  different  sense. 

Theorem  I.  Any  positive  number  may  he  added  to,  subtracted 
from,  or  midtiplied  by  both  numbers  of  an  inequality  without 
affecting  the  sense  of  the  inequality. 

Let  a>  b,  that  is,  let  a  —  b  =  k,  where  k  is  a  positive  number. 
If  m  is  a  positive  number,  evidently 


a  ±  m  —  (b  ±  m)  =  k, 

or 

a  ±m>  b±m. 

Similarly, 

ma  —  mb  =  mk, 

or 

ma  >  mb. 

The  other  statements  of  the  theorem  are  proved  similarly. 

Corollary.    Terms  may  be  transposed  from  one  side  of  an 
inequality  to  the  other  as  in  the  case  of  equations. 

269 


270  ADVANCED  ALGEBRA 

Let  a>  b  -{-  G. 

Subtract  c  from  both  sides  of  the  inequality  and  we  obtain  by 
Theorem  I 

a  —  c>  h. 

Theorem  IL  If  the  signs  of  both  sides  of  an  inequality  are 
changed,  the  sense  of  the  inequality  must  be  reversed,  that  is,  the 
>  sign  must  be  changed  to  <,  or  conversely. 

Let  a>  b,  that  is,  let  a  —  b  =  k,  where  ^  is  a  positive  number. 

Then  -a-\-b=-k, 

or  (—  «)—(—&)  =  —  7c, 

that  is,  by  definition,  —  a  < —  b. 

EXERCISES 

Prove  that  the  following  identities  are  tnie  for  all  real  positive  values  of 

the  letters. 

1.  a2  +  62  >  2  ah. 

Solution :  (a  —  h)^  is  always  positive. 

Thus  a2  -  2  a6  +  62  =  a2  +  62  _  2  a6  is  positive. 

That  is,  a2  +  62  >  2  ah. 

2.  3(a3  +  63)>a26  +  a62. 

3.  a2  +  62  +  c2  >  a6  +  ac  +  he. 

4.  (6  +  c)  (c  +  a)  (a  +  6)  >  8  ahc. 

5.  (a  +  6  +  c) (pfi  +  62  +  c2) > 9a6c. 

6.  62c2  +  c2a2  +  a262  >  a6c  (a  +  6  +  c). 

7.  3(a8  4-  63  +  c8) > (a  +  6  +  c)  {ah  +  6c  +  ca). 

8.  V(x  +  xi)2  +  (y  +  2/i)2  <  Vx2  +  2/2  +  y/x^^  +  yi^. 

9.  If  a2  +  62  =  1,  x2  +  y2  =  1^  prove  that  ax-\-hy<  1. 

10.  (a  +  6  -  c)2  +  (a  +  c  -  6)2  +  (6  +  c  -  a)2  >  a6  +  6c  4-  ca. 

11.  Show  that  the  sum  of  any  positive  number  (except  1)  and  its  reciprocal 
is  greater  than  2. 

12.  Prove  that  the  arithmetical  mean  of  two  unequal  positive  numbers 
always  exceeds  their  geometrical  mean. 


INEQUALITIES  271 

226.  Conditional  linear  inequalities.   If  we  wish  to  find  the 
values  of  x  for  which 

ax  -\-  b  <  c,  (1) 

where  a,  h,  and  c  are  numbers  and  a  is  positive,  we  may  find 
such  values  by  carrying  out  a  process  similar  to  that  of  solving  a 
linear  equation  in  one  variable. 

By  the  corollary,  §  225,  we  have  from  (1) 

ax  <  c  —  b. 


By  Theorem  I,  §  225,        x  <- 


227.  Conditional  quadratic  inequalities.  We  have  already 
shown  in  §  116  that  the  quadratic  expression  ax^  -\- bx  -\-  c  is  posi- 
tive or  negative,  when  the  equation 

ax^ -{- bx -{-  G  =  0  (1) 

has  imaginary  or  equal  roots,  according  as  a  is  positive  or  nega- 
tive. If  the  equation  has  distinct  real  roots,  the  expression  is 
positive  or  negative  for  values  between  those  roots  according  as 
a  is  negative  or  positive.  This  we  may  express  in  tabular  form 
as  follows,  for  all  values  of  x  excepting  the  roots  of  (1),  for 
which  of  course  the  expression  vanishes. 


a 

62  _  4  ac 

dx^  -\-  bx  +  c 

+ 

-orO 

Always  + 

- 

-orO 

Always  — 

+ 

+ 

-  for  X  between  roots,  +  for  other  values 

- 

+ 

+  for  X  between  roots,  -  for  other  values 

This  enables  us  to  answer  immediately  questions  like  the 
following : 

Example.  For  what  values  of  cc  is  —  2  ic^+a;  >  —  3  ?  By  the  corollary,  §  225, 
this  is  equivalent  to  the  question,  For  what  value  ofxis  —2x^-^x-\-S>0? 

Here  62  —  4  ac  =  1  -}-  24  =  25  is  positive.  The  roots  of  the  equation 
—  2  a;2  4-  a;  +  3  =  0  are  ic  =  —  1,  a;  =  f .  Thus  by  our  table  this  expression 
is  positive  for  all  values  of  x  between  —  1  and  |. 


272  ADVANCED  ALGEBRA 

EXERCISES 

For  what  values  of  x  are  the  following  inequalities  valid  ? 

1.  2x-3>0. 

3.   _x-l>7. 

-    9x      4     ^ 
3       7 

7.  .12x.+  .3<1.3. 

9.  3<5x-2. 

11.  x2-8x  +  22>6. 
13.  2x2-3x>5. 
15.  2x2 -4x<  -2. 
17.   -3x2  +  2x<2. 
19.  5x2-8x<l. 
21.  3x2>3x-3. 


2. 

4x-7>l. 

4. 

-  3  X  +  8  <  3. 

6. 

3   ^4^5 

8. 

3-4x>2. 

10. 

.s<i|-. 

12. 

x2  +  3x-2>l. 

14. 

-3x2-4x>8. 

16. 

3x2-9x>-6. 

18. 

-  x2  +  6  X  >  9. 

20. 

x2  <  X  -  1. 

22. 

3x>2x2-4. 

CHAPTEE  XXIII 

VARIATION 

228.  General  principles.  The  number  x  is  said  to  vary  directly 
as  the  number  y  when  the  ratio  of  ic  to  2/  is  constant. ,  This  we 
symbolize  by 

i»  oc  2/,  or  ^  =  A^,  (1) 

where  A;  is  a  constant. 

Thus  if  a  man  walks  at  a  uniform  speed,  the  distance  that  he 
goes  varies  directly  as  the  time.  If  the  length  of  the  altitude  of 
a  triangle  is  given,  the  area  of  the  triangle  varies  directly  as  the 
base.  The  volume  of  a  sphere  varies  directly  as  the  cube  of 
its  radius. 

The  number  x  is  said  to  vary  inversely  as  the  number  y  when 
X  varies  directly  as  the  reciprocal  of  y.  Thus  x  varies  inversely 
as  y  when 

05  oc  -5    OY  -  —  xy  =^  Uy  (2) 

y       i 
y 

where  A:  is  a  constant.  Thus  the  speed  of  a  horse  might  vary 
inversely  as  the  weight  of  his  load.  The  length  of  time  to  do 
a  piece  of  work  might  vary  inversely  as  the  nmnber  of  laborers 
employed. 

The  intensity  of  a  light  varies  inversely  as  the  square  of  the 
distance  of  the  light  from  the  point  of  observation.  If  I  repre- 
sents the  intensity  of  light  and  d  the  distance  of  the  light  from 
the  point  of  observation,  we  have 

Zoc|.   or|  =  ZcZ^  =  A5,  (3) 

where  A;  is  a  constant.  - '  - 

273 


274 


ADVANCED  ALGEBRA 


The  number  x  is  said  to  vary  jointly  as  y  and  z  when  it  varies 
directly  as  the  product  of  y  and  z.  Thus  x  varies  jointly  as  y 
and  z  when 


X  QC  yz,  or 


yz 


h 


(4) 


where  Aj  is  a  constant. 

Thus  a  man's  wages  might  vary  jointly  as  the  number  of  days 
and  the  number  of  hours  per  day  that  he  worked. 

The  number  x  is  said  to  vary  directly  as  y  and  inversely  as  z 

when  it  varies  directly  with  -•    Thus  the  force  of  the  attraction 

of  gravitation  between  two  bodies  varies  directly  as  their  masses 
and  inversely  as  the  squares  of  their  distances.  If  m  represents 
the  masses  of  two  bodies,  d  their  distance,  and  G  the  force  of 
their  attraction  due  to  gravity,  then 


m 
Goc^,  or 


=  k. 


(P) 


EXERCISES 

1.  If  a  varies  inversely  as  the  square  of  6,  and  if  a  =  2  when  6  =  3,  what 
is  the  value  of  a  when  6  is  18  ? 

Solution:   By  (3),  a62  =  k. 

We  can  determine  k  by  substituting  a  =  2,  6  =  3. 

2  .  9  =  A;. 
18  =  A:. 


Then 


a .  (18)2  =  18, 


or  a  =  tV- 

2.  The  volume  of  a  sphere  varies  as  the  cube  of  its  radius.  A  sphere  of 
radius  1  has  a  volume  4.19.    What  is  the  volume  of  a  sphere  of  radius  3  ? 

Solution:  Let  F  represent  the  volume  and  r  the  radius  of  the  sphere. 
Then  by  (1), 

\  =  k. 
4.19 


Determine  k  by  substituting. 


Then 


=  k. 


A;  =  4. 19. 

V       V 

--  =  —  =  4.19. 
(3)8      27 

F=  113.13. 


VARIATION  275 

3.  ltxyxx  +  y,  and  a;  =  1  when  y  =  1,  find  x  when  y  =  8. 

4.  The  area  of  a  circle  varies  as  the  square  of  the  radius.  If  a  circle  of 
radius  1  has  an  area  3.14,  find  the  area  of  a  circle  whose  radius  is  21. 

5.  Find  the  volume  of  a  sphere  whose  radius  is  .2. 
Hint.  See  exercise  2. 

6.  The  volume  of  a  circular  cylinder  varies  jointly  with  the  altitude  and 
the  square  of  the  radius  of  the  base.  A  cylinder  whose  altitude  and  radius 
are  each  1  has  a  volume  of  3. 14.  Find  the  volume  of  a  cylinder  whose 
altitude  is  16  and  whose  radius  is  3. 

7.  The  weight  of  a  body  of  a  given  material  varies  directly  with  its 
volume.  If  a  sphere  of  radius  1  inch  weighs  f  of  a  pound,  how  much  would 
a  ball  of  the  same  material  weigh  whose  radius  is  16  inches  ? 

8.  The  distance  fallen  by  an  object  starting  from  rest  varies  as  the  square 
of  the  time  of  falling.  If  a  body  falls  16  feet  in  1  second,  how  far  will  it 
fall  in  6  seconds  ? 

9.  A  body  falls  from  the  top  to  the  bottom  of  a  cliff  in  3|  seconds.  How 
high  is  the  cliff  ? 

10.  A  triangle  varies  in  area  jointly  as  its  base  and  altitude.  The  area 
of  a  triangle  whose  base  and  altitude  are  each  1  is  ^.  What  is  the  area  of  a 
triangle  whose  base  is  16  and  altitude  7? 

11.  If  6  men  do  a  piece  of  work  in  10  days,  how  long  will  it  take  5  men 
to  do  it? 

12.  If  3  men  working  8  hours  a  day  can  finish  a  piece  of  work  in  10  days, 
how  many  days  will  8  men  require  if  they  work  9  hours  a  day  ? 

13.  An  object  is  30  feet  from  a  light.  To  what  point  must  it  be  moved  in 
order  to  receive  (a)  half  as  much  light,  (b)  three  times  as  much  light  ? 

14.  The  weights  of  objects  near  the  earth  vary  inversely  as  the  squares 
of  their  distances  from  the  center  of  the  earth.  The  radius  of  the  earth  is 
about  4000  miles.  If  an  object  weighs  150  pounds  on  the  surface  of  the 
earth,  how  much  would  it  weigh  6000  miles  distant  from  the  center  ? 


CHAPTEE  XXIV 
PROBABILITY 

229.  Illustration.  If  a  bag  contains  3  wlaite  balls  and  4  blact 
balls,  and  1  ball  is  taken  out  at  random,  what  is  the  chance  that 
the  ball  drawn  will  be  white  ? 

This  question  we  may  answer  as  follows :  There  are  7  balls  in 
the  bag  and  we  are  as  likely  to  get  one  as  another.  Thus  a  ball 
may  be  drawn  in  7  different  ways.  Of  these  7  possible  ways  3 
will  produce  a  white  ball.  Thus  the  chance  that  the  ball  drawn 
will  be  white  is  3  to  7,  or  f.  The  chance  that  a  black  ball  will 
be  drawn  is  f . 

230.  General  statement.  It  is  plain  that  we  may  generalize 
this  illustration  as  follows :  If  an  event  may  happen  in  ^  ways 
and  fail  in  q  ways,  each  way  being  equally  probable,  the  chance 
or  probability  that  it  will  happen  in  one  of  the  ^  ways  is 


V 


The  chance  that  it  will  fail  is 


p  +  q 


(1) 


(2) 


The  sum  of  the  chances  of  the  event's  happening  and  failing 
is  1,  as  we  observe  by  adding  (1)  and  (2). 

The  odds  in  favor  of  the  event  are  the  ratio  of  the  chance  of 
happening  to  the  chance  of  failure.  In  this  case  the  odds  in 
favor  are 

I.  (3) 

q 

The  odds  against  the  event  are  -. 

P 
276 


PKOBABILITi  277 


EXERCISES 

1.  K  the  chance  of  an  event's  happening  is  ^^^  what  are  the  odds  in  its 
favor  ? 

v         1 
Solution :  By  (1),  -^—  =  — . 

Hence  lOjp  =  p  +  g, 

or  9p  =  g, 

«i      1 
or  -  =  - ,  which  by  (3)  are  the  odds  in  favor. 

q      9 

2.  From  a  pack  of  52  cards  3  are  missing.  What  is  the  chance  that  they 
are  all  of  a  particular  suit  P 

Solution :  The  number  of  combinations  of  62  cards  taken  3  at  a  time  is 

C52, 3  = — '    This  represents  p  ■{■  q.  The  number  of  combinations  of  the 

1  *  2  •  3  13  •  12  •  11 

13  cards  of  any  one  suit  taken  3  at  a  time  is  C13, 3  =         ^   „    •    This  repre- 
sentep, 

13  •  12  •  11 

^  p  1-2.3         13.12-11         11  11 

Thus  — =- —  = = =  ■ = . 

p-\-q      62 » 61  •  60      62.61-60      17-60      860 

1-2-3 

3.  What  is  the  chance  of  throwing  one  and  only  one  6  in  a  single  throw 
of  two  dice  ? 

Solution :  There  are  36  possible  ways  for  the  two  dice  to  fall.  This  repre- 
sents p  +  g.  Since  a  throw  of  two  sixes  is  excluded  there  are  6  throws  in 
which  each  die  would  be  a  6,  that  is,  10  in  all  in  which  a  6  appears.  This 
represents  p. 

P         10       6 
Thus  — - —  =  —  =  — 

p  +  q      36      18 

4.  A  bag  contains  8  white  and  12  black  balls.  What  is  the  chance  that  a 
ball  drawn  shall  be  (a)  white,  (b)  black  ? 

5.  A  bag  contains  4  red,  8  black,  and  12  white  balls.  What  is  the  chance 
that  a  ball  drawn  shall  be  (a)  red,  (b)  white,  (c)  not  black  ? 

6.  In  the  previous  problem,  if  3  balls  are  drawn,  what  is  the  chance  that 
(a)  all  are  black,  (b)  2  red  and  1  white  ? 

7.  What  is  the  chance  of  throwing  neither  a  3  nor  a  4  in  a  single  throw 
of  one  die  ? 

8.  What  is  the  chance  in  drawing  a  card  from  a  pack  that  it  be  (a)  an 
ace,  (b)  a  diamond,  (c)  a  face  card  ? 


278  ADVANCED  ALGEBRA 

9.  Three  cards  are  missing  from  a  pack.    What  is  the  chance  that  they 
are  (a)  of  one  color,  (b)  face  cards,  (c)  aces  ? 

10.  A   coin  is  tossed  twice.   What  is  the  chance  that  heads  will  fall 
once? 

11.  The  chance  that  an  event  will  happen  is  f.    What  are  the  odds  in  its 
favor  ? 

12.  The  odds  against  the  occurrence  of  an  event  are  |.    What  is  the  chance 
of  its  happening  ? 

13.'  What  is  the  chance  of  throwing  10  with  a  single  throw  of  two  dice  ? 

14.  A  squad  of  10  men  stand  in  line.    What  is  the  chance  that  A  and  B 
are  next  each  other  ? 

15.  What  is  the  chance  that  in  a  game  of  whist  a  player  has  6  trumps  ? 

16.  What  is  the  chance  that  in  a  game  of  whist  a  player  holds  4  aces  ? 


CHAPTER  XXV 
SCALES  OF  NOTATION 

231.  General  statement.    The  ordinary  numbers  with  which 
we  are  acquainted  are  expressed  by  means  of  powers  of  10.    Thus 
263  =  2 .  10^  +  6 .  10^  +  3. 
This  is  the  common  scale  of  notation,  and  10  is  called  the  radix 
of  the  scale. 

In  a  similar  manner  a  number  might  be  expressed  in  any  scale 
with  any  radix  other  than  10.    If  we  take  6  as  the  radix,  we  shall 
have  as  a  'number  in  this  scale,  for  instance, 
543  =  5  •  62  +  4  •  6  4-  3. 
In  this  scale  we  need  only  0  and  five  digits  to  express  every 
positive  integer. 

In  general,  if  r  is  the  radix  of  a  scale  of  notation,  any  positive 
integer  N  will  be  denoted  in  this  scale  as  follows : 

N=ao7^  +  a^r--^  +  a^r^-^  +  . . .  +  o,^.  (1) 

Theorem.    Any  positive .  integer  may  he  expressed  in  a  scale 
of  notation  of  radix  r. 

Suppose  we  have  a  positive  integer  N.    Let  ?•"  be  the  highest 
power  of  r  that  is  contained  in  N.    Then 

N  =  aoT-  +  iV'i, 
where  N^  is  less  than  r".    Suppose  that  on  dividing  iVj  by  r"-^  we 
obtain  ^^  ^  ^^^.„_i  ^  ^^^ 

where  N^  is  less  than  r"~\ 

Then  N  =  a^r^  +  air«-i  +  N^. 

Proceeding  in  this  manner  we  obtain  finally 

N  =  aor""  -h  0^1^""^  H h  a„, 

where  the  a's  are  positive  integers  less  than  r,  or  perhaps  zeros. 

279 


280  ADVANCED  ALGEBRA 

One  observes  that  the  symbol  10  indicates  the  radix  in  any 
system.  In  this  general  scale  we  need  a  0  and  r  —  1  digits  to 
express  every  possible  number. 

232.  Fundamental  operations.  In  the  four  fundamental  operar 
tions  in  the  common  scale  we  carry  and  borrow  10  in  computing. 
In  computing  in  a  scale  of  radix  6,  for  instance,  we  should  carry 
and  borrow  6.    If  the  radix  were  r,  we  should  carry  or  borrow  r. 

Thus  let  r  =  6.  Then  4  +  5=1-6+3=13.  Similarly,  5-3=2. 6  +  3  =  23. 
This  is  precisely  analogous  to  our  computation  in  the  common  scale,  where,  for 
instance,  we  would  have  9  +  8  =  1  •  10  +  7  =  17,  or  6  •  7  =  4  - 10  +  2  =  42. 

EXERCISES 

Perform  the  following  operations. 

1.  2361  +  4253  +  2140  ;  r  =  7. 

2361 
4253 
2140  - 


12114 


In  this  process,  since  3  +  1=4  and  is  less  than  the  radix,  there  is  nothing  to  carry. 
The  next  column  gives  6  +  5  +  4=  15  =2-7+1,  hence  we  write  down  1  and  carry  2.  The 
next  column  gives  3  +  2+1  +  2=8=1-7  +  1,  hence  we  write  down  1  and  carry  1.  Finally 
we  get  2  +  4  +  2  +  1  =  9=1-7  +  2,  hence  we  write  12. 

2.  4602-3714;  r  =  8. 

4602 
3714 


Since  we  cannot  take  4  from  2  we  horrow  one  from  the  next  place.  Since  the  radix 
is  8  this  amounts  to  8  units  in  the  first  place.  We  then  subtract  4  from  8  +  2,  which 
leaves  6.  In  borrowing  1  from  0  that  digit  is  really  reduced  to  7  and  the  preceding  digit 
to  5 ;  then  subtracting  1  from  7  we  get  6.  Since  we  cannot  take  7  from  5  we  borrow  8 
again  and  take  7  from  5  +  8=  13,  which  leaves  6.  Since  1  has  been  borrowed  from  the  4 
we  see  the  subtraction  is  complete  since  3-3=0. 

3.  4321 .  432 ;  r  =  6. 

4821 

432 

14142 

24013 

33334 

4143222 

In  multiplying  by  2  we  have  nothing  to  carry  until  we  multiply  3  by  2.  This  gives 
6=1-5  +  1.  Hence  we  put  down  1  and  carry  1  to  the  product  of  2  and  4.  The  addition  of 
the  partial  products  is  carried  out  as  in  exercise  1. 


SCALES  OF  NOTATION  281 

4.  32130  H-  43 ;  r  =  6. 

43 1 32130 1 430 
300 
213 
213 
00 

In  making  an  estimate  for  the  first  figure  in  the  quotient  we  divide  32  by  4,  keeping 
in  mind  that  for  this  purpose  32=3-64-2.  Thus  4  is  contained  in  20  just  5  times,  but 
since  our  entire  divisor  is  43  we  take  4  as  the  first  figure  in  the  quotient.  The  multipli- 
cations are  of  course  performed  as  in  exercise  3,  excepting  that  here  6  is  the  radix. 

5.  4361  +  2635  +  5542  ;  r  =  1.  6.   5344  -  3456  ;  r  =  7. 
7.  2340  .  4101 ;  r=6.                            8.  6435  •  35  ;  r  =  7. 

9.  2003455  -  403 ;  r  =  6.  10.  344032  -  321 ;  r  =  5. 

11.  534401  -  443524 ;  r  =  6.  12.  425  +  254  +  542  +  452  ;  r=6. 

233.  Change  of  scale.  If  we  have  a  number  in  the  scale  of 
radix  r,  we  may  find  the  expression  for  that  number  in  the  com- 
mon scale  by  writing  the  number  in  form  (1),  §  231,  and  carrying 
out  the  indicated  operations.. 

Example.    Convert  4635,  where  r  =  7,  into  the  ordinary  scale. 
4635  =  4- 73 +  6.  72 +  3-7+5 
=  4  •  343  +  6  .  40  +  3  .  7  +  5 
=  1692. 

If  we  have  a  number  in  the  common  scale,  we  may  express  it 
in  the  scale  with  radix  r  as  follows  :  If  the  number  is  N,  we  have 
to  determine  the  integers  aQ,  a^,  •   • ,  a„  in  the  expression 

N  =  ao7^  +  «^i^""^  H h  a^-iT  +  ««•  (1) 

Divide  (1)  by  r.    We  have 

:^=  aor"-i  +  a,r-'  H-  a^,_,  +  -  =  iV'  +-; 
r  r  r 

that  is,  the  remainder  a^  of  this  division  is  the  last  digit  in  the 
expression  desired. 

Divide  N'  by  r  and  we  obtain 

—  =  JSJ-"  =  a^r^-^  +  a^r^-^  H -|-  ^^^^ ; 

r  r 


282  ADVANCED  ALGEBRA 

that  is,  the  remainder  from  this  division  is  the  next  to  the  last 
digit  in  the  desired  expression.  Proceeding  in  this  way  we  obtain 
all  the  digits  fl^„,  c^„  _  i,  •  •  • ,  ^i,  «o- 

Example.    Express  37496  in  the  scale  with  radix  7. 

7)  37496 

7)  5356  remainder  4 

1)  765  remainder  1 

•  7)  109  remainder  2 

7)  15  remainder  4 

7)2  remainder  1 

0  remainder  2 

The  number  in  scale  r  =  7  is  214214. 

To  change  a  number  from  any  scale  rj  to  any  other  scale  r^j 
we  may  first  change  the  number  to  the  scale  of  10  and  then  by 
the  process  just  given  to  the  scale  r^.  The  process  indicated  in 
the  preceding  example  may  be  employed  directly  to  change  from 
any  scale  to  any  other,  provided  the  division  is  carried  out  in  the 
scale  in  which  the  number  is  given.  One  of  these  methods  may 
be  used  to  check  the  other. 

Example.    Change  34503  from  scale  r  =  6  to  one  in  which  r  =  9. 

34503  =  3.  6* +  4- 63 +5.  62 +  3  =  4935  in  scale  of  10.      ♦ 

9)4935 

9)  548  remainder  3 

9)60  remainder  8 

9)6  remainder  6 

0  remainder  6 

Thus  34503  in  scale  of  6  becomes  6683  in  scale  of  9. 

Check:  9)34503 

9)2312  remainder  3  ^"  carrying  out  this  division  it  must 

■  X  •    A       Q  ^®  kept  in  mind  that  the  dividends  are  in 

92140  remamder  8  g^^jg  ^f  g^  ^l^jle  ^^Q  remainders  are  to  be 

9)  10  remainder  6  in  scale  of  9. 
0  remainder  6 

234.  Fractions.  In  the  ordinary  notation  we  express  fractional 
numbers  by  digits  following  the  decimal  point.  This  notation 
may  also  be  used  in  a  scale  with  any  radix. 


SCALES  OF  NOTATION  283 

Thus  the  expression  .5421  stands  for 

10      102  ^  10«      10* 
in  the  common  scale. 

In  the  scale  with  radix  r  it  stands  for 

ry%  ry%**  n%9  niT 

The  process  of  changing  the  scale  for  fractions  is  performed 
in  accordance  with  the  same  principles  as  are  employed  in  the 
change  of  scale  for  integers.  The  following  examples  suffice  to 
illustrate  it. 

Example  1.    Express  .5421  in  the  scale  of  6  as  a  decimal  fraction. 

.6421  =  5  +  1+1+1 
6      62      68      64 

5.63  +  4.62+2-6+1      1237       ^,,, 
6*  1296 

Example  2.    Express  .439  as  a  fraction  for  radix  6. 

Let  .439  =  ^  +  A  +  l  +  l  +  .... 

6      62      68      6* 

Multiplying  by  6,  2.634  =  a  +  ^  +  ^  +  ^  +  • . . . 

b      o-'      6' 

Thus  a  =  2  and  we  have  .634  =  -  +  — f- 1 . 

6  ^  62      68 

c       d 
Multiplying  by  6,  3.804  =  6  +  -  +  —  +  ... . 

6      62 

c       d 
Thus  6  =  3  and  we  have  .804  =  -  +  —  +  ••. . 

6      62^ 

Multiplying  by  6,  4.824  =  c +  -  +  •• .. 

6 

d 
Thus  c  =  4  and  we  have  .824  =  -  +  .... 

6 

Multiplying  by  (T,  4.944  =  d  +  • . .. 

Thus  d  =  4. 

The  fraction  in  scale  of  radix  6  is  then  .2344  •  •  «„ 


284  ADVANCED  ALGEBRA 

EXERCISES 

1.  Express  the  following  as  decimal  fractions, 
(a)  .374;  r  =  8.  (b)  .4352;  r  =  6. 
(c)  .2231 ;  r  =  4.                                    (d)  .2001 ;  r  =  3. 

2.  Express  the  decimal  fraction  .296  as  a  radix  fraction  for  r  =  5. 

3.  Express  the  decimal  fraction  .3405  as  a  radix  fraction  for  r  =  Q. 

34 

4.  Express  —  as  a  radix  fraction  for  r  =  4. 

^        128 

5.  Express as  a  radix  fraction  for  r  =  5. 

^        626 

6.  In  what  scale  is  42  expressed  as  1120? 

Solution :   We  seek  r  where 

r3  +  r2  +  2  r  =  42. 

This  is  equivalent  to  finding  a  positive  integral  root  of  the  equation 

r3  +  r2  +  2  r  -  42  =  0. 
By  synthetic  division, 

1  +  1  +  2  -  42(2 
+  2  +  6  +  16 
+  3  4-  8-26 
1  +  1+  2-4213 
+  3  +  12  +  42 
+  4+14  0 
Thus  3  is  the  value  sought. 

Check :   3^  +  32  +  2  •  3  =  27  +  9  +  6  =  42. 

7.  In  what  scale  is  2704  denoted  by  20304  ? 

8.  In  what  scale  is  256  denoted  by  10000  ? 

9.  In  what  scale  is  .1664  denoted  by  .0404  ? 
10.  Show  that  1331  is  a  perfect  cube. 

235.  Duodecimals.  We  may  apply  some  of  the  foregoing 
processes  to  mensuration.  If  we  take  one  foot  as  a  unit  and  the 
radix  as  12,  we  may  express  distances  in  the  so-called  duodecimal 
notation.  Thus  2  ft.  6  in.  is  represented  in  the  duodecimal  scale 
by  2.6.  Since  in  a  scale  of  radix  r  we  need  ?•  —  1  symbols,  we 
will  let  10  =  ^  and  11  =  e.  Thus  21  ft.  10  in.  would  be  expressed 
in  duodecimal  notation  as  19.^.  We  may  now  find  areas  and  vol- 
umes in  this  notation  much  more  readily  than  by  the  usual  method 
of  converting  all  distances  to  inches. 


SCALES  OF  NOTATION  285 

Example.  Multiply  8  ft.  3  in.  by  3  ft.  10  in.    We  multiply  8  •  3  by  3.i 

8.3  in  the  duodecimal  scale.    To  convert  the  result  to  square 

3-t  feet  and  square  inches  we  must  keep  in  mind  that  27.76 

6t6  ^  2  .  12  +  7  +  tV  +  xf^  =  31  sq.   ft.   90   sq.  in.,  since  144 

— —  square  inches  equal  one  square  foot. 

This  example  suggests  the  following  method  of  multiplying 
distances : 

EuLE.  Express  the  distances  in  duodecimal  notation  with  the 
foot  as  a  unit. 

Multiply  in  the  scale  for  which  r  =  12. 

In  the  product  change  the  part  on  the  left  of  the  point  from 
duodecimal  to  decimal  scale. 

Multiply  the  digit  following  the  point  hy  12^  and  add  to  the 
last  figure  to  obtain  the  square  inches  in  the  result. 

EXERCISES 

1.  Multiply  the  following  : 

(a)  13  ft.  4  in.  by  67  ft.  11  in. 

Solution:  11.4 

57.e 
1028 
794 

568 


636.68  =  905  sq.  ft.  80  sq.  in. 

(b)  10  ft.  6  in.  by  12  ft.  2  in. 

(c)  8  ft.  4  in.  by  11  ft.  11  in. 

(d)  23  ft.  6  in.  by  47  ft.  8  in. 

(e)  41  ft.  6  in.  by  36  ft.  1  in. 

2.  What  is  the  area  of  a  room  16  ft.  2  in.  by  10  ft.  3  in.  ? 

3.  What  is  the  area  of  a  walk  60  ft.  6  in.  by  3  ft.  3  in.  ? 

4.  What  is  the  area  of  a  city  lot  62  ft.  6  in.  by  163  ft.  7  in.  ? 


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